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Calculus of Parametric Curves
Determine derivatives and equations of tangents for parametric curves.
Derivatives of Parametric Equations
We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations
\[x(t)=2t+3,\ y(t)=3t-4,\ -2\le t\le 3.\]The graph of this curve appears in . It is a line segment starting at \((-1,-10)\) and ending at \((9,5).\)
We can eliminate the parameter by first solving the equation \(x(t)=2t+3\) for t:
\[\begin{array}{lll}x(t) & = & 2t+3 \\ x-3 & = & 2t \\ t & = & \frac{x-3}{2}.\end{array}\]Substituting this into \(y(t),\) we obtain
\[\begin{array}{lll}y(t) & = & 3t-4 \\ y & = & 3(\frac{x-3}{2})-4 \\ y & = & \frac{3x}{2}-\frac{9}{2}-4 \\ y & = & \frac{3x}{2}-\frac{17}{2}.\end{array}\]The slope of this line is given by \(\frac{dy}{dx}=\frac{3}{2}.\) Next we calculate \({x}^{'}(t)\) and \({y}^{'}(t).\) This gives \({x}^{'}(t)=2\) and \({y}^{'}(t)=3.\) Notice that \(\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt}=\frac{3}{2}.\) This is no coincidence, as outlined in the following theorem.
Condensed — the full section is in OpenStax Calculus Volume 3.
Second-Order Derivatives
Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function \(y=f(x)\) is defined to be the derivative of the first derivative; that is,
\[\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}[\frac{dy}{dx}].\]Since \(\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt},\) we can replace the \(y\) on both sides of this equation with \(\frac{dy}{dx}.\) This gives us
\[\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}(\frac{dy}{dx})=\frac{(d\text{/}dt)(dy\text{/}dx)}{dx\text{/}dt}.\]If we know \(dy\text{/}dx\) as a function of t, then this formula is straightforward to apply.
Example
Try it.
Calculate the second derivative \({d}^{2}y\text{/}d{x}^{2}\) for the plane curve defined by the parametric equations \(x(t)={t}^{2}-3,y(t)=2t-1,-3\le t\le 4.\)
Solution
From we know that \(\frac{dy}{dx}=\frac{2}{2t}=\frac{1}{t}.\) Using , we obtain
\[\frac{{d}^{2}y}{d{x}^{2}}=\frac{(d\text{/}dt)(dy\text{/}dx)}{dx\text{/}dt}=\frac{(d\text{/}dt)(1\text{/}t)}{2t}=\frac{\text{-}{t}^{-2}}{2t}=-\frac{1}{2{t}^{3}}.\]Integrals Involving Parametric Equations
Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations \(x(t)=t-\text{sin}\ t,\ y(t)=1-\text{cos}\ t.\) Suppose we want to find the area of the shaded region in the following graph.
To derive a formula for the area under the curve defined by the functions
\[x=x(t),\ y=y(t),\ a\le t\le b,\]we assume that \(x(t)\) is increasing on the interval \(t\in [a,b]\) and \(x(t)\) is differentiable and start with an equal partition of the interval \(a\le t\le b.\) Suppose \({t}_{0}=a<{t}_{1}<{t}_{2}<\text{\cdots }<{t}_{n}=b\) and consider the following graph.
We use rectangles to approximate the area under the curve. The height of the \(i\)th rectangle is \(y({t}_{i-1})\), so an approximation to the area is
\[\begin{array}{llll}\begin{array}{ll} & \sum _{i=1}^{n}y({t}_{i-1})(x({t}_{i})-x({t}_{i-1})) \\ = & \sum _{i=1}^{n}y({t}_{i-1})\frac{(x({t}_{i})-x({t}_{i-1}))}{({t}_{i}-{t}_{i-1})}({t}_{i}-{t}_{i-1}) \\ & \to {\int }_{a}^{b}y(t)x'(t)dt\text{as max}\{({t}_{i}-{t}_{i-1})\}\to 0\end{array}\end{array}\]This follows from results obtained in Calculus 1 for the function \(y({t}_{i-1})\frac{(x({t}_{i})-x({t}_{i-1}))}{({t}_{i}-{t}_{i-1})}.\)
Then a Riemann sum for the area is
\[{A}_{n}=\sum _{i=1}^{n}y(\overset{—}{{t}_{i}})\ (x({t}_{i})-x({t}_{i-1})).\]Multiplying and dividing each area by \({t}_{i}-{t}_{i-1}\) gives
\[{A}_{n}=\sum _{i=1}^{n}y(x({\overset{-}{t}}_{i}))\ (\frac{x({t}_{i})-x({t}_{i-1})}{{t}_{i}-{t}_{i-1}})({t}_{i}-{t}_{i-1})=\sum _{i=1}^{n}y(x({\overset{-}{t}}_{i}))\ (\frac{x({t}_{i})-x({t}_{i-1})}{\text{\Delta }t})\text{\Delta }t.\]\[A=\underset{n\to \infty }{\text{lim}}{A}_{n}={\int }_{a}^{b}y(t){x}^{'}(t)\ dt.\]Example
Try it.
Find the area under the curve of the cycloid defined by the equations
\[x(t)=t-\text{sin}\ t,\ y(t)=1-\text{cos}\ t,\ 0\le t\le 2\pi .\]Solution
Using , we have
\[\begin{array}{ll}A & ={\int }_{a}^{b}y(t){x}^{'}(t)\ dt \\ & ={\int }_{0}^{2\pi }(1-\text{cos}\ t)(1-\text{cos}\ t)\ dt \\ & ={\int }_{0}^{2\pi }(1-2\ \text{cos}\ t+{\text{cos}}^{2}t)dt \\ & ={\int }_{0}^{2\pi }(1-2\ \text{cos}\ t+\frac{1+\text{cos}\ 2t}{2})\ dt \\ & ={\int }_{0}^{2\pi }(\frac{3}{2}-2\ \text{cos}\ t+\frac{\text{cos}\ 2t}{2})\ dt \\ & ={\frac{3t}{2}-2\ \text{sin}\ t+\frac{\text{sin}\ 2t}{4}|}_{0}^{2\pi } \\ & =3\pi .\end{array}\]Condensed — the full section is in OpenStax Calculus Volume 3.
Arc Length of a Parametric Curve
In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point A to point B along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.
Given a plane curve defined by the functions \(x=x(t),y=y(t),a\le t\le b,\) we start by partitioning the interval \([a,b]\) into n equal subintervals: \({t}_{0}=a<{t}_{1}<{t}_{2}<\text{\cdots }<{t}_{n}=b.\) The width of each subinterval is given by \(\text{\Delta }t=(b-a)\text{/}n.\) We can calculate the length of each line segment:
\[\begin{array}{l} \\ {d}_{1}=\sqrt{{(x({t}_{1})-x({t}_{0}))}^{2}+{(y({t}_{1})-y({t}_{0}))}^{2}} \\ {d}_{2}=\sqrt{{(x({t}_{2})-x({t}_{1}))}^{2}+{(y({t}_{2})-y({t}_{1}))}^{2}}\ \text{etc}.\end{array}\]Then add these up. We let s denote the exact arc length and \({s}_{n}\) denote the approximation by n line segments:
\[s\approx \sum _{k=1}^{n}{s}_{k}=\sum _{k=1}^{n}\sqrt{{(x({t}_{k})-x({t}_{k-1}))}^{2}+{(y({t}_{k})-y({t}_{k-1}))}^{2}}.\]If we assume that \(x(t)\) and \(y(t)\) are differentiable functions of t, then the Mean Value Theorem (Introduction to the Applications of Derivatives) applies, so in each subinterval \([{t}_{k-1},{t}_{k}]\) there exist \({\overset{\wedge}{t}}_{k}\) and \({\overset{˜}{t}}_{k}\) such that
\[\begin{array}{l} \\ x({t}_{k})-x({t}_{k-1})={x}^{'}({\overset{\wedge}{t}}_{k})({t}_{k}-{t}_{k-1})={x}^{'}({\overset{\wedge}{t}}_{k})\text{\Delta }t \\ y({t}_{k})-y({t}_{k-1})={y}^{'}({\overset{˜}{t}}_{k})({t}_{k}-{t}_{k-1})={y}^{'}({\overset{˜}{t}}_{k})\text{\Delta }t.\end{array}\]Therefore becomes
\[\begin{array}{ll}s & \approx \sum _{k=1}^{n}{s}_{k} \\ & =\sum _{k=1}^{n}\sqrt{{({x}^{'}({\overset{\wedge}{t}}_{k})\text{\Delta }t)}^{2}+{({y}^{'}({\overset{˜}{t}}_{k})\text{\Delta }t)}^{2}} \\ & =\sum _{k=1}^{n}\sqrt{{({x}^{'}({\overset{\wedge}{t}}_{k}))}^{2}{(\text{\Delta }t)}^{2}+{({y}^{'}({\overset{˜}{t}}_{k}))}^{2}{(\text{\Delta }t)}^{2}} \\ & =(\sum _{k=1}^{n}\sqrt{{({x}^{'}({\overset{\wedge}{t}}_{k}))}^{2}+{({y}^{'}({\overset{˜}{t}}_{k}))}^{2}})\text{\Delta }t.\end{array}\]This is a Riemann sum that approximates the arc length over a partition of the interval \([a,b].\) If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives
When taking the limit, the values of \({\overset{\wedge}{t}}_{k}\) and \({\overset{˜}{t}}_{k}\) are both contained within the same ever-shrinking interval of width \(\text{\Delta }t,\) so they must converge to the same value.
Condensed — the full section is in OpenStax Calculus Volume 3.
Surface Area Generated by a Parametric Curve
Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function \(y=f(x)\) from \(x=a\) to \(x=b,\) revolved around the x-axis:
\[S=2\pi {\int }_{a}^{b}f(x)\sqrt{1+{({f}^{'}(x))}^{2}}dx.\]We now consider a volume of revolution generated by revolving a parametrically defined curve \(x=x(t),y=y(t),a\le t\le b\) around the x-axis as shown in the following figure.
The analogous formula for a parametrically defined curve is
\[S=2\pi {\int }_{a}^{b}y(t)\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt\]provided that \(y(t)\) is not negative on \([a,b].\)
Example
Try it.
Find the surface area of a sphere of radius r centered at the origin.
Solution
We start with the curve defined by the equations
\[x(t)=r\ \text{cos}\ t,\ y(t)=r\ \text{sin}\ t,\ 0\le t\le \pi .\]This generates an upper semicircle of radius r centered at the origin as shown in the following graph.
When this curve is revolved around the x-axis, it generates a sphere of radius r. To calculate the surface area of the sphere, we use :
\[\begin{array}{ll}S & =2\pi {\int }_{a}^{b}y(t)\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt \\ & =2\pi {\int }_{0}^{\pi }r\ \text{sin}\ t\sqrt{{(\text{-}r\ \text{sin}\ t)}^{2}+{(r\ \text{cos}\ t)}^{2}}dt \\ & =2\pi {\int }_{0}^{\pi }r\ \text{sin}\ t\sqrt{{r}^{2}{\text{sin}}^{2}t+{r}^{2}{\text{cos}}^{2}t}\ dt \\ & =2\pi {\int }_{0}^{\pi }r\ \text{sin}\ t\sqrt{{r}^{2}({\text{sin}}^{2}t+{\text{cos}}^{2}t)}dt \\ & =2\pi {\int }_{0}^{\pi }{r}^{2}\text{sin}\ t\ dt \\ & =2\pi {r}^{2}({\text{-}\text{cos}\ t|}_{0}^{\pi }) \\ & =2\pi {r}^{2}(\text{-}\text{cos}\ \pi +\text{cos}\ 0) \\ & =4\pi {r}^{2}.\end{array}\]This is, in fact, the formula for the surface area of a sphere.
Key Concepts
- The derivative of the parametrically defined curve \(x=x(t)\) and \(y=y(t)\) can be calculated using the formula \(\frac{dy}{dx}=\frac{{y}^{'}(t)}{{x}^{'}(t)}.\) Using the derivative, we can find the equation of a tangent line to a parametric curve.
- The area between a parametric curve and the x-axis can be determined by using the formula \(A={\int }_{{t}_{1}}^{{t}_{2}}y(t){x}^{'}(t)\ dt.\)
- The arc length of a parametric curve can be calculated by using the formula \(s={\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{(\frac{dx}{dt})}^{2}+{(\frac{dy}{dt})}^{2}}dt.\)
- The surface area of a volume of revolution revolved around the x-axis is given by \(S=2\pi {\int }_{a}^{b}y(t)\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt.\) If the curve is revolved around the y-axis, then the formula is \(S=2\pi {\int }_{a}^{b}x(t)\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt.\)
Key Equations
| Derivative of parametric equations | \(\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt}=\frac{{y}^{'}(t)}{{x}^{'}(t)}\) |
| Second-order derivative of parametric equations | \(\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}(\frac{dy}{dx})=\frac{(d\text{/}dt)(dy\text{/}dx)}{dx\text{/}dt}\) |
| Area under a parametric curve | \(A={\int }_{a}^{b}y(t){x}^{'}(t)\ dt\) |
| Arc length of a parametric curve | \(s={\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{(\frac{dx}{dt})}^{2}+{(\frac{dy}{dt})}^{2}}dt\) |
| Surface area generated by a parametric curve | \(S=2\pi {\int }_{a}^{b}y(t)\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt\) |
Calculus of Parametric Curves
For the following exercises, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.
For the following exercises, determine the slope of the tangent line, then find an equation of the tangent line at the given value of the parameter.
For the following exercises, find all points on the curve that have the given slope.
For the following exercises, write the equation of the tangent line in Cartesian coordinates for the given parameter t.
For the following exercises, find \({d}^{2}y\text{/}d{x}^{2}.\)
For the following exercises, find points on the curve at which tangent line is horizontal or vertical.
For the following exercises, find \(dy\text{/}dx\) at the value of the parameter.
Condensed — the full section is in OpenStax Calculus Volume 3.
Derivatives of Parametric Equations
We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations
\[x(t)=2t+3,\ y(t)=3t-4,\ -2\le t\le 3.\]The graph of this curve appears in . It is a line segment starting at \((-1,-10)\) and ending at \((9,5).\)
We can eliminate the parameter by first solving the equation \(x(t)=2t+3\) for t:
\[\begin{array}{lll}x(t) & = & 2t+3 \\ x-3 & = & 2t \\ t & = & \frac{x-3}{2}.\end{array}\]Substituting this into \(y(t),\) we obtain
\[\begin{array}{lll}y(t) & = & 3t-4 \\ y & = & 3(\frac{x-3}{2})-4 \\ y & = & \frac{3x}{2}-\frac{9}{2}-4 \\ y & = & \frac{3x}{2}-\frac{17}{2}.\end{array}\]The slope of this line is given by \(\frac{dy}{dx}=\frac{3}{2}.\) Next we calculate \({x}^{'}(t)\) and \({y}^{'}(t).\) This gives \({x}^{'}(t)=2\) and \({y}^{'}(t)=3.\) Notice that \(\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt}=\frac{3}{2}.\) This is no coincidence, as outlined in the following theorem.
Condensed — the full section is in OpenStax Calculus Volume 2.
Second-Order Derivatives
Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function \(y=f(x)\) is defined to be the derivative of the first derivative; that is,
\[\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}[\frac{dy}{dx}].\]Since \(\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt},\) we can replace the \(y\) on both sides of this equation with \(\frac{dy}{dx}.\) This gives us
\[\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}(\frac{dy}{dx})=\frac{(d\text{/}dt)(dy\text{/}dx)}{dx\text{/}dt}.\]If we know \(dy\text{/}dx\) as a function of t, then this formula is straightforward to apply.
Example
Try it.
Calculate the second derivative \({d}^{2}y\text{/}d{x}^{2}\) for the plane curve defined by the parametric equations \(x(t)={t}^{2}-3,y(t)=2t-1,-3\le t\le 4.\)
Solution
From we know that \(\frac{dy}{dx}=\frac{2}{2t}=\frac{1}{t}.\) Using , we obtain
\[\frac{{d}^{2}y}{d{x}^{2}}=\frac{(d\text{/}dt)(dy\text{/}dx)}{dx\text{/}dt}=\frac{(d\text{/}dt)(1\text{/}t)}{2t}=\frac{\text{-}{t}^{-2}}{2t}=-\frac{1}{2{t}^{3}}.\]Integrals Involving Parametric Equations
Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations \(x(t)=t-\text{sin}\ t,\ y(t)=1-\text{cos}\ t.\) Suppose we want to find the area of the shaded region in the following graph.
To derive a formula for the area under the curve defined by the functions
\[x=x(t),\ y=y(t),\ a\le t\le b,\]we assume that \(x(t)\) is increasing on the interval \(t\in [a,b]\) and \(x(t)\) is differentiable and start with an equal partition of the interval \(a\le t\le b.\) Suppose \({t}_{0}=a<{t}_{1}<{t}_{2}<\text{\cdots }<{t}_{n}=b\) and consider the following graph.
We use rectangles to approximate the area under the curve. The height of the \(i\)th rectangle is \(y({t}_{i-1})\), so an approximation to the area is
\[\begin{array}{llll}\begin{array}{ll} & \sum _{i=1}^{n}y({t}_{i-1})(x({t}_{i})-x({t}_{i-1})) \\ = & \sum _{i=1}^{n}y({t}_{i-1})\frac{(x({t}_{i})-x({t}_{i-1}))}{({t}_{i}-{t}_{i-1})}({t}_{i}-{t}_{i-1}) \\ & \to {\int }_{a}^{b}y(t)x'(t)dt\text{as max}\{({t}_{i}-{t}_{i-1})\}\to 0\end{array}\end{array}\]This follows from results obtained in Calculus 1 for the function \(y({t}_{i-1})\frac{(x({t}_{i})-x({t}_{i-1}))}{({t}_{i}-{t}_{i-1})}.\)
Then a Riemann sum for the area is
\[{A}_{n}=\sum _{i=1}^{n}y(\overset{—}{{t}_{i}})\ (x({t}_{i})-x({t}_{i-1})).\]Multiplying and dividing each area by \({t}_{i}-{t}_{i-1}\) gives
\[{A}_{n}=\sum _{i=1}^{n}y(x({\overset{-}{t}}_{i}))\ (\frac{x({t}_{i})-x({t}_{i-1})}{{t}_{i}-{t}_{i-1}})({t}_{i}-{t}_{i-1})=\sum _{i=1}^{n}y(x({\overset{-}{t}}_{i}))\ (\frac{x({t}_{i})-x({t}_{i-1})}{\text{\Delta }t})\text{\Delta }t.\]\[A=\underset{n\to \infty }{\text{lim}}{A}_{n}={\int }_{a}^{b}y(t){x}^{'}(t)\ dt.\]Example
Try it.
Find the area under the curve of the cycloid defined by the equations
\[x(t)=t-\text{sin}\ t,\ y(t)=1-\text{cos}\ t,\ 0\le t\le 2\pi .\]Solution
Using , we have
\[\begin{array}{ll}A & ={\int }_{a}^{b}y(t){x}^{'}(t)\ dt \\ & ={\int }_{0}^{2\pi }(1-\text{cos}\ t)(1-\text{cos}\ t)\ dt \\ & ={\int }_{0}^{2\pi }(1-2\ \text{cos}\ t+{\text{cos}}^{2}t)dt \\ & ={\int }_{0}^{2\pi }(1-2\ \text{cos}\ t+\frac{1+\text{cos}\ 2t}{2})\ dt \\ & ={\int }_{0}^{2\pi }(\frac{3}{2}-2\ \text{cos}\ t+\frac{\text{cos}\ 2t}{2})\ dt \\ & ={\frac{3t}{2}-2\ \text{sin}\ t+\frac{\text{sin}\ 2t}{4}|}_{0}^{2\pi } \\ & =3\pi .\end{array}\]Condensed — the full section is in OpenStax Calculus Volume 2.
Arc Length of a Parametric Curve
In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point A to point B along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.
Given a plane curve defined by the functions \(x=x(t),y=y(t),a\le t\le b,\) we start by partitioning the interval \([a,b]\) into n equal subintervals: \({t}_{0}=a<{t}_{1}<{t}_{2}<\text{\cdots }<{t}_{n}=b.\) The width of each subinterval is given by \(\text{\Delta }t=(b-a)\text{/}n.\) We can calculate the length of each line segment:
\[\begin{array}{l} \\ {d}_{1}=\sqrt{{(x({t}_{1})-x({t}_{0}))}^{2}+{(y({t}_{1})-y({t}_{0}))}^{2}} \\ {d}_{2}=\sqrt{{(x({t}_{2})-x({t}_{1}))}^{2}+{(y({t}_{2})-y({t}_{1}))}^{2}}\ \text{etc}.\end{array}\]Then add these up. We let s denote the exact arc length and \({s}_{n}\) denote the approximation by n line segments:
\[s\approx \sum _{k=1}^{n}{s}_{k}=\sum _{k=1}^{n}\sqrt{{(x({t}_{k})-x({t}_{k-1}))}^{2}+{(y({t}_{k})-y({t}_{k-1}))}^{2}}.\]If we assume that \(x(t)\) and \(y(t)\) are differentiable functions of t, then the Mean Value Theorem (Introduction to the Applications of Derivatives) applies, so in each subinterval \([{t}_{k-1},{t}_{k}]\) there exist \({\overset{\wedge}{t}}_{k}\) and \({\overset{˜}{t}}_{k}\) such that
\[\begin{array}{l} \\ x({t}_{k})-x({t}_{k-1})={x}^{'}({\overset{\wedge}{t}}_{k})({t}_{k}-{t}_{k-1})={x}^{'}({\overset{\wedge}{t}}_{k})\text{\Delta }t \\ y({t}_{k})-y({t}_{k-1})={y}^{'}({\overset{˜}{t}}_{k})({t}_{k}-{t}_{k-1})={y}^{'}({\overset{˜}{t}}_{k})\text{\Delta }t.\end{array}\]Therefore becomes
\[\begin{array}{ll}s & \approx \sum _{k=1}^{n}{s}_{k} \\ & =\sum _{k=1}^{n}\sqrt{{({x}^{'}({\overset{\wedge}{t}}_{k})\text{\Delta }t)}^{2}+{({y}^{'}({\overset{˜}{t}}_{k})\text{\Delta }t)}^{2}} \\ & =\sum _{k=1}^{n}\sqrt{{({x}^{'}({\overset{\wedge}{t}}_{k}))}^{2}{(\text{\Delta }t)}^{2}+{({y}^{'}({\overset{˜}{t}}_{k}))}^{2}{(\text{\Delta }t)}^{2}} \\ & =(\sum _{k=1}^{n}\sqrt{{({x}^{'}({\overset{\wedge}{t}}_{k}))}^{2}+{({y}^{'}({\overset{˜}{t}}_{k}))}^{2}})\text{\Delta }t.\end{array}\]This is a Riemann sum that approximates the arc length over a partition of the interval \([a,b].\) If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives
When taking the limit, the values of \({\overset{\wedge}{t}}_{k}\) and \({\overset{˜}{t}}_{k}\) are both contained within the same ever-shrinking interval of width \(\text{\Delta }t,\) so they must converge to the same value.
Condensed — the full section is in OpenStax Calculus Volume 2.
Surface Area Generated by a Parametric Curve
Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function \(y=f(x)\) from \(x=a\) to \(x=b,\) revolved around the x-axis:
\[S=2\pi {\int }_{a}^{b}f(x)\sqrt{1+{({f}^{'}(x))}^{2}}dx.\]We now consider a volume of revolution generated by revolving a parametrically defined curve \(x=x(t),y=y(t),a\le t\le b\) around the x-axis as shown in the following figure.
The analogous formula for a parametrically defined curve is
\[S=2\pi {\int }_{a}^{b}y(t)\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt\]provided that \(y(t)\) is not negative on \([a,b].\)
Example
Try it.
Find the surface area of a sphere of radius r centered at the origin.
Solution
We start with the curve defined by the equations
\[x(t)=r\ \text{cos}\ t,\ y(t)=r\ \text{sin}\ t,\ 0\le t\le \pi .\]This generates an upper semicircle of radius r centered at the origin as shown in the following graph.
When this curve is revolved around the x-axis, it generates a sphere of radius r. To calculate the surface area of the sphere, we use :
\[\begin{array}{ll}S & =2\pi {\int }_{a}^{b}y(t)\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt \\ & =2\pi {\int }_{0}^{\pi }r\ \text{sin}\ t\sqrt{{(\text{-}r\ \text{sin}\ t)}^{2}+{(r\ \text{cos}\ t)}^{2}}dt \\ & =2\pi {\int }_{0}^{\pi }r\ \text{sin}\ t\sqrt{{r}^{2}{\text{sin}}^{2}t+{r}^{2}{\text{cos}}^{2}t}\ dt \\ & =2\pi {\int }_{0}^{\pi }r\ \text{sin}\ t\sqrt{{r}^{2}({\text{sin}}^{2}t+{\text{cos}}^{2}t)}dt \\ & =2\pi {\int }_{0}^{\pi }{r}^{2}\text{sin}\ t\ dt \\ & =2\pi {r}^{2}({\text{-}\text{cos}\ t|}_{0}^{\pi }) \\ & =2\pi {r}^{2}(\text{-}\text{cos}\ \pi +\text{cos}\ 0) \\ & =4\pi {r}^{2}.\end{array}\]This is, in fact, the formula for the surface area of a sphere.
Key Concepts
- The derivative of the parametrically defined curve \(x=x(t)\) and \(y=y(t)\) can be calculated using the formula \(\frac{dy}{dx}=\frac{{y}^{'}(t)}{{x}^{'}(t)}.\) Using the derivative, we can find the equation of a tangent line to a parametric curve.
- The area between a parametric curve and the x-axis can be determined by using the formula \(A={\int }_{{t}_{1}}^{{t}_{2}}y(t){x}^{'}(t)\ dt.\)
- The arc length of a parametric curve can be calculated by using the formula \(s={\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{(\frac{dx}{dt})}^{2}+{(\frac{dy}{dt})}^{2}}dt.\)
- The surface area of a volume of revolution revolved around the x-axis is given by \(S=2\pi {\int }_{a}^{b}y(t)\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt.\) If the curve is revolved around the y-axis, then the formula is \(S=2\pi {\int }_{a}^{b}x(t)\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt.\)
Key Equations
| Derivative of parametric equations | \(\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt}=\frac{{y}^{'}(t)}{{x}^{'}(t)}\) |
| Second-order derivative of parametric equations | \(\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}(\frac{dy}{dx})=\frac{(d\text{/}dt)(dy\text{/}dx)}{dx\text{/}dt}\) |
| Area under a parametric curve | \(A={\int }_{a}^{b}y(t){x}^{'}(t)\ dt\) |
| Arc length of a parametric curve | \(s={\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{(\frac{dx}{dt})}^{2}+{(\frac{dy}{dt})}^{2}}dt\) |
| Surface area generated by a parametric curve | \(S=2\pi {\int }_{a}^{b}y(t)\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt\) |
Calculus of Parametric Curves
For the following exercises, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.
For the following exercises, determine the slope of the tangent line, then find an equation of the tangent line at the given value of the parameter.
For the following exercises, find all points on the curve that have the given slope.
For the following exercises, write the equation of the tangent line in Cartesian coordinates for the given parameter t.
For the following exercises, find \({d}^{2}y\text{/}d{x}^{2}.\)
For the following exercises, find points on the curve at which tangent line is horizontal or vertical.
For the following exercises, find \(dy\text{/}dx\) at the value of the parameter.
Condensed — the full section is in OpenStax Calculus Volume 2.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Calculate the derivative \(\frac{dy}{dx}\) for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs.
- \(x(t)={t}^{2}-3,\ y(t)=2t-1,\ -3\le t\le 4\)
- \(x(t)=2t+1,\ y(t)={t}^{3}-3t+4,\ -2\le t\le 2\)
- \(x(t)=5\ \text{cos}\ t,\ y(t)=5\ \text{sin}\ t,\ 0\le t\le 2\pi\)
Разкрийте отговора
- To apply , first calculate \({x}^{'}(t)\) and \({y}^{'}(t)\text{:}\)
\[\begin{array}{l}{x}^{'}(t)=2t \\ {y}^{'}(t)=2.\end{array}\]
Next substitute these into the equation:
\[\begin{array}{l}\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt} \\ \frac{dy}{dx}=\frac{2}{2t} \\ \frac{dy}{dx}=\frac{1}{t}.\end{array}\]
This derivative is undefined when \(t=0.\) Calculating \(x(0)\) and \(y(0)\) gives \(x(0)={(0)}^{2}-3=-3\) and \(y(0)=2(0)-1=-1,\) which corresponds to the point \((-3,-1)\) on the graph. The graph of this curve is a parabola opening to the right, and the point \((-3,-1)\) is its vertex as shown.
- To apply , first calculate \({x}^{'}(t)\) and \({y}^{'}(t)\text{:}\)
\[\begin{array}{l}{x}^{'}(t)=2 \\ {y}^{'}(t)=3{t}^{2}-3.\end{array}\]
Next substitute these into the equation:
\[\begin{array}{l}\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt} \\ \frac{dy}{dx}=\frac{3{t}^{2}-3}{2}.\end{array}\]
This derivative is zero when \(t=\pm 1.\) When \(t=-1\) we have
\[x(-1)=2(-1)+1=-1\ \text{and}\ y(-1)={(-1)}^{3}-3(-1)+4=-1+3+4=6,\]
which corresponds to the point \((-1,6)\) on the graph. When \(t=1\) we have
\[x(1)=2(1)+1=3\ \text{and}\ y(1)={(1)}^{3}-3(1)+4=1-3+4=2,\]
which corresponds to the point \((3,2)\) on the graph. The point \((3,2)\) is a relative minimum and the point \((-1,6)\) is a relative maximum, as seen in the following graph.
- To apply , first calculate \({x}^{'}(t)\) and \({y}^{'}(t)\text{:}\)
\[\begin{array}{l}{x}^{'}(t)=-5\ \text{sin}\ t \\ {y}^{'}(t)=5\ \text{cos}\ t.\end{array}\]
Next substitute these into the equation:
\[\begin{array}{l}\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt} \\ \frac{dy}{dx}=\frac{5\ \text{cos}\ t}{-5\ \text{sin}\ t} \\ \frac{dy}{dx}=\text{-}\text{cot}\ t.\end{array}\]
This derivative is zero when \(\text{cos}\ t=0\) and is undefined when \(\text{sin}\ t=0.\) This gives \(t=0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},\text{and}\ 2\pi\) as critical points for t. Substituting each of these into \(x(t)\) and \(y(t),\) we obtain
\(t\) \(x(t)\) \(y(t)\) 0 5 0 \(\frac{\pi }{2}\) 0 5 \(\pi\) −5 0 \(\frac{3\pi }{2}\) 0 −5 \(2\pi\) 5 0
These points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero.
-
Calculate the derivative \(dy\text{/}dx\) for the plane curve defined by the equations
\[x(t)={t}^{2}-4t,\ y(t)=2{t}^{3}-6t,\ -2\le t\le 3\]and locate any critical points on its graph.
Разкрийте отговора
\({x}^{'}(t)=2t-4\) and \({y}^{'}(t)=6{t}^{2}-6,\) so \(\frac{dy}{dx}=\frac{6{t}^{2}-6}{2t-4}=\frac{3{t}^{2}-3}{t-2}.\)
This expression is undefined when \(t=2\) and equal to zero when \(t=\pm 1.\)
-
Find an equation of the tangent line to the curve defined by the equations
\[x(t)={t}^{2}-3,\ y(t)=2t-1,\ -3\le t\le 4\ \text{when}\ t=2.\]Разкрийте отговора
First find the slope of the tangent line using , which means calculating \({x}^{'}(t)\) and \({y}^{'}(t)\text{:}\)
\[\begin{array}{l}{x}^{'}(t)=2t \\ {y}^{'}(t)=2.\end{array}\]Next substitute these into the equation:
\[\begin{array}{l}\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt} \\ \frac{dy}{dx}=\frac{2}{2t} \\ \frac{dy}{dx}=\frac{1}{t}.\end{array}\]When \(t=2,\) \(\frac{dy}{dx}=\frac{1}{2},\) so this is the slope of the tangent line. Calculating \(x(2)\) and \(y(2)\) gives
\[x(2)={(2)}^{2}-3=1\ \text{and}\ y(2)=2(2)-1=3,\]which corresponds to the point \((1,3)\) on the graph (). Now use the point-slope form of the equation of a line to find an equation of the tangent line:
\[\begin{array}{lll}y-{y}_{0} & = & m(x-{x}_{0}) \\ y-3 & = & \frac{1}{2}(x-1) \\ y-3 & = & \frac{1}{2}x-\frac{1}{2} \\ y & = & \frac{1}{2}x+\frac{5}{2}.\end{array}\] -
Find an equation of the tangent line to the curve defined by the equations
\[x(t)={t}^{2}-4t,\ y(t)=2{t}^{3}-6t,\ -2\le t\le 10\ \text{when}\ t=5.\]Разкрийте отговора
The equation of the tangent line is \(y=24x+100.\)
-
Calculate the second derivative \({d}^{2}y\text{/}d{x}^{2}\) for the plane curve defined by the parametric equations \(x(t)={t}^{2}-3,y(t)=2t-1,-3\le t\le 4.\)
Разкрийте отговора
From we know that \(\frac{dy}{dx}=\frac{2}{2t}=\frac{1}{t}.\) Using , we obtain
\[\frac{{d}^{2}y}{d{x}^{2}}=\frac{(d\text{/}dt)(dy\text{/}dx)}{dx\text{/}dt}=\frac{(d\text{/}dt)(1\text{/}t)}{2t}=\frac{\text{-}{t}^{-2}}{2t}=-\frac{1}{2{t}^{3}}.\] -
Calculate the second derivative \({d}^{2}y\text{/}d{x}^{2}\) for the plane curve defined by the equations
\[x(t)={t}^{2}-4t,\ y(t)=2{t}^{3}-6t,\ -2\le t\le 3\]and locate any critical points on its graph.
Разкрийте отговора
\(\frac{{d}^{2}y}{d{x}^{2}}=\frac{3{t}^{2}-12t+3}{2{(t-2)}^{3}}.\) Critical points \((5,4),(-3,-4),\text{and}\ (-4,4).\)
-
Find the area under the curve of the cycloid defined by the equations
\[x(t)=t-\text{sin}\ t,\ y(t)=1-\text{cos}\ t,\ 0\le t\le 2\pi .\]Разкрийте отговора
Using , we have
\[\begin{array}{ll}A & ={\int }_{a}^{b}y(t){x}^{'}(t)\ dt \\ & ={\int }_{0}^{2\pi }(1-\text{cos}\ t)(1-\text{cos}\ t)\ dt \\ & ={\int }_{0}^{2\pi }(1-2\ \text{cos}\ t+{\text{cos}}^{2}t)dt \\ & ={\int }_{0}^{2\pi }(1-2\ \text{cos}\ t+\frac{1+\text{cos}\ 2t}{2})\ dt \\ & ={\int }_{0}^{2\pi }(\frac{3}{2}-2\ \text{cos}\ t+\frac{\text{cos}\ 2t}{2})\ dt \\ & ={\frac{3t}{2}-2\ \text{sin}\ t+\frac{\text{sin}\ 2t}{4}|}_{0}^{2\pi } \\ & =3\pi .\end{array}\] -
Find the area under the curve of the hypocycloid defined by the equations
\[x(t)=3\ \text{cos}\ t+\text{cos}\ 3t,\ y(t)=3\ \text{sin}\ t-\text{sin}\ 3t,\ 0\le t\le \pi .\]Разкрийте отговора
\(A=3\pi\) (Note that the integral formula actually yields a negative answer. This is due to the fact that \(x(t)\) is a decreasing function over the interval \([0,2\pi ];\) that is, the curve is traced from right to left.)
-
Find the arc length of the semicircle defined by the equations
\[x(t)=3\ \text{cos}\ t,\ y(t)=3\ \text{sin}\ t,\ 0\le t\le \pi .\]Разкрийте отговора
The values \(t=0\) to \(t=\pi\) trace out the red curve in . To determine its length, use :
\[\begin{array}{ll}s & ={\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{(\frac{dx}{dt})}^{2}+{(\frac{dy}{dt})}^{2}}dt \\ & ={\int }_{0}^{\pi }\sqrt{{(-3\ \text{sin}\ t)}^{2}+{(3\ \text{cos}\ t)}^{2}}dt \\ & ={\int }_{0}^{\pi }\sqrt{9\ {\text{sin}}^{2}t+9\ {\text{cos}}^{2}t}\ dt \\ & ={\int }_{0}^{\pi }\sqrt{9({\text{sin}}^{2}t+{\text{cos}}^{2}t)}dt \\ & ={\int }_{0}^{\pi }3dt={3t|}_{0}^{\pi }=3\pi .\end{array}\]Note that the formula for the arc length of a semicircle is \(\pi r\) and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.
-
Find the arc length of the curve defined by the equations
\[x(t)=3{t}^{2},\ y(t)=2{t}^{3},\ 1\le t\le 3.\]Разкрийте отговора
\(s=2({10}^{3\text{/}2}-{2}^{3\text{/}2})\approx 57.589\)
-
Find the surface area of a sphere of radius r centered at the origin.
Разкрийте отговора
We start with the curve defined by the equations
\[x(t)=r\ \text{cos}\ t,\ y(t)=r\ \text{sin}\ t,\ 0\le t\le \pi .\]This generates an upper semicircle of radius r centered at the origin as shown in the following graph.
When this curve is revolved around the x-axis, it generates a sphere of radius r. To calculate the surface area of the sphere, we use :
\[\begin{array}{ll}S & =2\pi {\int }_{a}^{b}y(t)\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}}dt \\ & =2\pi {\int }_{0}^{\pi }r\ \text{sin}\ t\sqrt{{(\text{-}r\ \text{sin}\ t)}^{2}+{(r\ \text{cos}\ t)}^{2}}dt \\ & =2\pi {\int }_{0}^{\pi }r\ \text{sin}\ t\sqrt{{r}^{2}{\text{sin}}^{2}t+{r}^{2}{\text{cos}}^{2}t}\ dt \\ & =2\pi {\int }_{0}^{\pi }r\ \text{sin}\ t\sqrt{{r}^{2}({\text{sin}}^{2}t+{\text{cos}}^{2}t)}dt \\ & =2\pi {\int }_{0}^{\pi }{r}^{2}\text{sin}\ t\ dt \\ & =2\pi {r}^{2}({\text{-}\text{cos}\ t|}_{0}^{\pi }) \\ & =2\pi {r}^{2}(\text{-}\text{cos}\ \pi +\text{cos}\ 0) \\ & =4\pi {r}^{2}.\end{array}\]This is, in fact, the formula for the surface area of a sphere.
-
Find the surface area generated when the plane curve defined by the equations
\[x(t)={t}^{3},\ y(t)={t}^{2},\ 0\le t\le 1\]is revolved around the x-axis.
Разкрийте отговора
\(A=\frac{\pi (494\sqrt{13}+128)}{1215}\)
-
\(\begin{array}{ll}x=3+t, & y=1-t\end{array}\)
-
\(\begin{array}{ll}x=8+2t, & y=1\end{array}\)
Разкрийте отговора
0
-
\(\begin{array}{ll}x=4-3t, & y=-2+6t\end{array}\)
-
\(\begin{array}{ll}x=-5t+7, & y=3t-1\end{array}\)
Разкрийте отговора
\(\frac{-3}{5}\)
-
\(\begin{array}{ll}x=3\ \text{sin}\ t, & y=3\ \text{cos}\ t,\ t=\frac{\pi }{4}\end{array}\)
-
\(\begin{array}{ll}x=\text{cos}\ t, & y=8\ \text{sin}\ t,\end{array}t=\frac{\pi }{2}\)
Разкрийте отговора
\(\text{Slope}=0;\) \(y=8.\)
-
\(\begin{array}{ll}x=2t, & y={t}^{3},\ t=-1\end{array}\)
-
\(\begin{array}{ll}x=t+\frac{1}{t}, & y=t-\frac{1}{t},\ t=1\end{array}\)
Разкрийте отговора
Slope is undefined; \(x=2.\)
-
\(\begin{array}{ll}x=\sqrt{t}, & y=2t,\ t=4\end{array}\)
-
\(\begin{array}{ll}x=4\ \text{cos}\ t, & y=4\ \text{sin}\ t,\end{array}\) slope = 0.5
Разкрийте отговора
\(\tan t=-2\ (\frac{4}{\sqrt{5}},\frac{-8}{\sqrt{5}}),(\frac{4}{\sqrt{5}},\frac{-8}{\sqrt{5}})\), \(\left(\frac{-4}{\sqrt{5}},\frac{8}{\sqrt{5}}\right)\).
-
\(\begin{array}{ll}x=2\ \text{cos}\ t, & y=8\ \text{sin}\ t,\ \text{slope}=-1\end{array}\)
-
\(\begin{array}{ll}x=t+\frac{1}{t}, & y=t-\frac{1}{t},\ \text{slope}=1\end{array}\)
Разкрийте отговора
No points possible; undefined expression.
-
\(\begin{array}{ll}x=2+\sqrt{t}, & y=2-4t,\ \text{slope}=0\end{array}\)
-
\(\begin{array}{ll}x={e}^{\sqrt{t}}, & y=1-\text{ln}\ {t}^{2},\ t=1\end{array}\)
Разкрийте отговора
\(y=\text{-}(\frac{4}{e})x+5\)
-
\(\begin{array}{ll}x=t\ \text{ln}\ t, & y={\text{sin}}^{2}t,\end{array}t=\frac{\pi }{4}\)
-
\(\begin{array}{ll}x={e}^{t}, & y={(t-1)}^{2},\ \text{at}(1,1)\end{array}\)
Разкрийте отговора
\(y=-2x+3\)
-
For \(x=\text{sin}(2t),y=2\ \text{sin}\ t\) where \(0\le t<2\pi .\) Find all values of t at which a horizontal tangent line exists.
-
For \(x=\text{sin}(2t),y=2\ \text{sin}\ t\) where \(0\le t<2\pi .\) Find all values of t at which a vertical tangent line exists.
Разкрийте отговора
\(\frac{\pi }{4},\frac{5\pi }{4},\frac{3\pi }{4},\frac{7\pi }{4}\)
-
Find all points on the curve \(x=4\ \text{sin}(t),y=4\ \text{cos}(t)\) that have the slope of \(0.5\)
-
Find \(\frac{dy}{dx}\) for \(x=\text{sin}(t),y=\text{cos}(t).\)
Разкрийте отговора
\(\frac{dy}{dx}=\text{-}\text{tan}(t)\)
-
Find an equation of the tangent line to \(x=\text{sin}(t),y=\text{cos}(t)\) at \(t=\frac{\pi }{4}.\)
-
For the curve \(x=4t,y=3t-2,\) find the slope and concavity of the curve at \(t=3.\)
Разкрийте отговора
\(\frac{dy}{dx}=\frac{3}{4}\) and \(\frac{{d}^{2}y}{d{x}^{2}}=0,\) so the curve is neither concave up nor concave down at \(t=3.\) Therefore the graph is linear and has a constant slope but no concavity.
-
For the parametric curve whose equation is \(x=4\ \text{cos}\ \theta ,y=4\ \text{sin}\ \theta ,\) find the slope and concavity of the curve at \(\theta =\frac{\pi }{4}.\)
-
Find the slope and concavity for the curve whose equation is \(x=2+\text{sec}\ \theta ,y=1+2\ \text{tan}\ \theta\) at \(\theta =\frac{\pi }{6}.\)
Разкрийте отговора
\(\frac{dy}{dx}=4,\frac{{d}^{2}y}{d{x}^{2}}=-6\sqrt{3};\) the curve is concave down at \(\theta =\frac{\pi }{6}.\)
-
Find all points on the curve \(x=t+4,y={t}^{3}-3t\) at which there are vertical and horizontal tangents.
-
Find all points on the curve \(x=\text{sec}\ \theta ,y=\text{tan}\ \theta\) at which horizontal and vertical tangents exist.
Разкрийте отговора
No horizontal tangents. Vertical tangents at \((1,0),(-1,0).\)
-
\(\begin{array}{ll}x={t}^{4}-1, & y=t-{t}^{2}\end{array}\)
-
\(\begin{array}{ll}x=\text{sin}(\pi t), & y=\text{cos}(\pi t)\end{array}\)
Разкрийте отговора
\(\text{-}{\text{sec}}^{3}(\pi t)\)
Symbols used here
Add a_k for k = 1 up to n.
Antiderivative (indefinite) or signed area from a to b (definite).
The non-negative number whose square (n-th power) is x.
Not a number: "grows without bound" in limits and intervals.
Ratio of a circle's circumference to its diameter, 3.14159…
x belongs to A; every element of A is in B.
Logical connectives.
Instantaneous rate of change; slope of the graph.
i² = −1.
Equal to the precision shown, not exactly.
Inequalities that allow equality; < and > exclude it.
The two sides are different.
Least upper bound, greatest lower bound.
Derivative with respect to x, holding the other variables fixed.
Vector of partial derivatives; points uphill.
Integral over a region of the plane; integral around a closed curve.
How to: Calculus of Parametric Curves
- Determine derivatives and equations of tangents for parametric curves.
- Find the area under a parametric curve.
- Use the equation for arc length of a parametric curve.
- Apply the formula for surface area to a volume generated by a parametric curve.
- To apply
- To apply
- To apply
- The derivative of the parametrically defined curve
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
Опитай си.
Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
Още в Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems