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Tangent Planes and Linear Approximations

Determine the equation of a plane tangent to a given surface at a point.

Tangent Planes

Intuitively, it seems clear that, in a plane, only one line can be tangent to a curve at a point. However, in three-dimensional space, many lines can be tangent to a given point. If these lines lie in the same plane, they determine the tangent plane at that point. A tangent plane at a regular point contains all of the lines tangent to that point. A more intuitive way to think of a tangent plane is to assume the surface is smooth at that point (no corners). Then, a tangent line to the surface at that point in any direction does not have any abrupt changes in slope because the direction changes smoothly.

For a tangent plane to a surface to exist at a point on that surface, it is sufficient for the function that defines the surface to be differentiable at that point, defined later in this section. We define the term tangent plane here and then explore the idea intuitively.

To see why this formula is correct, let’s first find two tangent lines to the surface \(S.\) The equation of the tangent line to the curve that is represented by the intersection of \(S\) with the vertical trace given by \(x={x}_{0}\) is \(z=f({x}_{0},{y}_{0})+{f}_{y}({x}_{0},{y}_{0})(y-{y}_{0}).\) Similarly, the equation of the tangent line to the curve that is represented by the intersection of \(S\) with the vertical trace given by \(y={y}_{0}\) is \(z=f({x}_{0},{y}_{0})+{f}_{x}({x}_{0},{y}_{0})(x-{x}_{0}).\) A parallel vector to the first tangent line is \(a=j+{f}_{y}({x}_{0},{y}_{0})k;\) a parallel vector to the second tangent line is \(b=i+{f}_{x}({x}_{0},{y}_{0})k.\) We can take the cross product of these two vectors:

\[\begin{array}{llllllll}a\ \times \ b & =(j+{f}_{y}({x}_{0},{y}_{0})k)\ \times \ (i+{f}_{x}({x}_{0},{y}_{0})k) \\ & =|\begin{array}{lll}i & j & k \\ 0 & 1 & {f}_{y}({x}_{0},{y}_{0}) \\ 1 & 0 & {f}_{x}({x}_{0},{y}_{0})\end{array}| \\ & ={f}_{x}({x}_{0},{y}_{0})i+{f}_{y}({x}_{0},{y}_{0})j-k.\end{array}\]

Solving this equation for \(z\) gives .

The graph of this function follows.

Condensed: the full section is in OpenStax Calculus Volume 3.

Linear Approximations

Recall from Linear Approximations and Differentials that the formula for the linear approximation of a function \(f(x)\) at the point \(x=a\) is given by

\[y\approx f(a)+f'(a)(x-a).\]

The diagram for the linear approximation of a function of one variable appears in the following graph.

The tangent line can be used as an approximation to the function \(f(x)\) for values of \(x\) reasonably close to \(x=a.\) When working with a function of two variables, the tangent line is replaced by a tangent plane, but the approximation idea is much the same.

Notice that this equation also represents the tangent plane to the surface defined by \(z=f(x,y)\) at the point \(({x}_{0},{y}_{0}).\) The idea behind using a linear approximation is that, if there is a point \(({x}_{0},{y}_{0})\) at which the precise value of \(f(x,y)\) is known, then for values of \((x,y)\) reasonably close to \(({x}_{0},{y}_{0}),\) the linear approximation (i.e., tangent plane) yields a value that is also reasonably close to the exact value of \(f(x,y)\) (). Furthermore the plane that is used to find the linear approximation is also the tangent plane to the surface at the point \(({x}_{0},{y}_{0}).\)

Condensed: the full section is in OpenStax Calculus Volume 3.

Differentiability

When working with a function \(y=f(x)\) of one variable, the function is said to be differentiable at a point \(x=a\) if \({f}^{'}(a)\) exists. Furthermore, if a function of one variable is differentiable at a point, the graph is “smooth” at that point (i.e., no corners exist) and a tangent line is well-defined at that point.

The idea behind differentiability of a function of two variables is connected to the idea of smoothness at that point. In this case, a surface is considered to be smooth at point \(P\) if a tangent plane to the surface exists at that point. If a function is differentiable at a point, then a tangent plane to the surface exists at that point. Recall the formula for a tangent plane at a point \(({x}_{0},{y}_{0})\) is given by

\[z=f({x}_{0},{y}_{0})+{f}_{x}({x}_{0},{y}_{0})(x-{x}_{0})+{f}_{y}({x}_{0},{y}_{0})(y-{y}_{0}),\]

For a tangent plane to exist at the point \(({x}_{0},{y}_{0}),\) the partial derivatives must therefore exist at that point. However, this is not a sufficient condition for smoothness, as was illustrated in . In that case, the partial derivatives existed at the origin, but the function also had a corner on the graph at the origin.

The last term in is referred to as the error term and it represents how closely the tangent plane comes to the surface in a small neighborhood \((\delta\) disk) of point \(P.\) For the function \(f\) to be differentiable at \(P,\) the function must be smooth, that is, the graph of \(f\) must be close to the tangent plane for points near \(P.\)

\[\begin{array}{lll}f(x,y) & = & f(0,0)+{f}_{x}(0,0)(x-0)+{f}_{y}(0,0)(y-0)+E(x,y) \\ E(x,y) & = & \frac{xy}{\sqrt{{x}^{2}+{y}^{2}}}.\end{array}\]

Condensed: the full section is in OpenStax Calculus Volume 3.

Differentials

In Linear Approximations and Differentials we first studied the concept of differentials. The differential of \(y,\) written \(dy,\) is defined as \({f}^{'}(x)dx.\) The differential is used to approximate \(\Delta y=f(x+\Delta x)-f(x),\) where \(\Delta x=dx.\) Extending this idea to the linear approximation of a function of two variables at the point \(({x}_{0},{y}_{0})\) yields the formula for the total differential for a function of two variables.

Notice that the symbol \(∂\) is not used to denote the total differential; rather, \(d\) appears in front of \(z.\) Now, let’s define \(\Delta z=f(x+\Delta x,y+\Delta y)-f(x,y).\) We use \(dz\) to approximate \(\Delta z,\) so

\[\Delta z\approx dz={f}_{x}({x}_{0},{y}_{0})dx+{f}_{y}({x}_{0},{y}_{0})dy.\]

Therefore, the differential is used to approximate the change in the function \(z=f({x}_{0},{y}_{0})\) at the point \(({x}_{0},{y}_{0})\) for given values of \(\Delta x\) and \(\Delta y.\) Since \(\Delta z=f(x+\Delta x,y+\Delta y)-f(x,y),\) this can be used further to approximate \(f(x+\Delta x,y+\Delta y)\text{:}\)

\[\begin{array}{ll}f(x+\Delta x,y+\Delta y) & =f(x,y)+\Delta z \\ & \approx f(x,y)+{f}_{x}({x}_{0},{y}_{0})\Delta x+{f}_{y}({x}_{0},{y}_{0})\Delta y.\end{array}\]

See the following figure.

One such application of this idea is to determine error propagation. For example, if we are manufacturing a gadget and are off by a certain amount in measuring a given quantity, the differential can be used to estimate the error in the total volume of the gadget.

Condensed: the full section is in OpenStax Calculus Volume 3.

Differentiability of a Function of Three Variables

All of the preceding results for differentiability of functions of two variables can be generalized to functions of three variables. First, the definition:

If a function of three variables is differentiable at a point \(({x}_{0},{y}_{0},{z}_{0}),\) then it is continuous there. Furthermore, continuity of first partial derivatives at that point guarantees differentiability.

Key Concepts

  • The analog of a tangent line to a curve is a tangent plane to a surface for functions of two variables.
  • Tangent planes can be used to approximate values of functions near known values.
  • A function is differentiable at a point if it is ”smooth” at that point (i.e., no corners or discontinuities exist at that point).
  • The total differential can be used to approximate the change in a function \(z=f({x}_{0},{y}_{0})\) at the point \(({x}_{0},{y}_{0})\) for given values of \(\Delta x\) and \(\Delta y.\)

Key Equations

Tangent plane\(z=f({x}_{0},{y}_{0})+{f}_{x}({x}_{0},{y}_{0})(x-{x}_{0})+{f}_{y}({x}_{0},{y}_{0})(y-{y}_{0})\)
Linear approximation\(L(x,y)=f({x}_{0},{y}_{0})+{f}_{x}({x}_{0},{y}_{0})(x-{x}_{0})+{f}_{y}({x}_{0},{y}_{0})(y-{y}_{0})\)
Total differential\(dz={f}_{x}({x}_{0},{y}_{0})dx+{f}_{y}({x}_{0},{y}_{0})dy.\)
Differentiability (two variables)\(f(x,y)=f({x}_{0},{y}_{0})+{f}_{x}({x}_{0},{y}_{0})(x-{x}_{0})+{f}_{y}({x}_{0},{y}_{0})(y-{y}_{0})+E(x,y),\)
where the error term \(E\) satisfies
\(\underset{(x,y)\to ({x}_{0},{y}_{0})}{\text{lim}}\frac{E(x,y)}{\sqrt{{(x-{x}_{0})}^{2}+{(y-{y}_{0})}^{2}}}=0.\)
Differentiability (three variables)\(\begin{array}{ll}f(x,y) & =f({x}_{0},{y}_{0},{z}_{0})+{f}_{x}({x}_{0},{y}_{0},{z}_{0})(x-{x}_{0})+{f}_{y}({x}_{0},{y}_{0},{z}_{0})(y-{y}_{0}) \\ & \ +{f}_{z}({x}_{0},{y}_{0},{z}_{0})(z-{z}_{0})+E(x,y,z),\end{array}\)
where the error term \(E\) satisfies
\(\underset{(x,y,z)\to ({x}_{0},{y}_{0},{z}_{0})}{\text{lim}}\frac{E(x,y,z)}{\sqrt{{(x-{x}_{0})}^{2}+{(y-{y}_{0})}^{2}+{(z-{z}_{0})}^{2}}}=0.\)

Tangent Planes and Linear Approximations

For the following exercises, find a unit normal vector to the surface at the indicated point.

For the following exercises, as a useful review for techniques used in this section, find a normal vector and a tangent vector at point \(P.\)

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (Hint: Solve for \(z\) in terms of \(x\) and \(y.)\)

For the following exercises, find parametric equations for the normal line to the surface at the indicated point. (Recall that to find the equation of a line in space, you need a point on the line, \({P}_{0}({x}_{0,}{y}_{0},{z}_{0}),\) and a vector \(n=〈a,b,c〉\) that is parallel to the line. Then the equation of the line is \(x-{x}_{0}=at,y-{y}_{0}=bt,z-{z}_{0}=ct.)\)

For the following exercises, use the figure shown here.

For the following exercises, complete each task.

For the following exercises, find the linear approximation of each function at the indicated point.

이제 너 계산기는 이것을 해결하지 않지만, 그 조각은 계산 가능합니다. 아래의 하나를 시도하거나 자신의 것을 입력하십시오.

연습 (40)

먼저 종이에 각각을 시도해 보세요. 확인하기 위해 답을 드러내세요. 확인된 답은 각 단계의 솔버에서 열 수 있습니다.

  1. Find an equation of the tangent plane to the surface defined by the function \(f(x,y)=2{x}^{2}-3xy+8{y}^{2}+2x-4y+4\) at point \((2,-1).\)

    답을 드러내세요

    First, we must calculate \({f}_{x}(x,y)\) and \({f}_{y}(x,y),\) then use with \({x}_{0}=2\) and \({y}_{0}=-1\text{:}\)

    \[\begin{array}{lll}{f}_{x}(x,y) & = & 4x-3y+2 \\ {f}_{y}(x,y) & = & -3x+16y-4 \\ f(2,-1) & = & 2{(2)}^{2}-3(2)(-1)+8{(-1)}^{2}+2(2)-4(-1)+4=34. \\ {f}_{x}(2,-1) & = & 4(2)-3(-1)+2=13 \\ {f}_{y}(2,-1) & = & -3(2)+16(-1)-4=-26.\end{array}\]

    Then becomes

    \[\begin{array}{l}z=f({x}_{0},{y}_{0})+{f}_{x}({x}_{0},{y}_{0})(x-{x}_{0})+{f}_{y}({x}_{0},{y}_{0})(y-{y}_{0}) \\ z=34+13(x-2)-26(y-(-1)) \\ z=34+13x-26-26y-26 \\ z=13x-26y-18.\end{array}\]

    (See the following figure).

  2. Find an equation of the tangent plane to the surface defined by the function \(f(x,y)={x}^{3}-{x}^{2}y+{y}^{2}-2x+3y-2\) at point \((-1,3).\)

    답을 드러내세요

    \(z=7x+8y-3\)

  3. Find an equation of the tangent plane to the surface defined by the function \(f(x,y)=\text{sin}(2x)\text{cos}(3y)\) at the point \((\pi \text{/}3,\pi \text{/}4).\)

    답을 드러내세요

    First, calculate \({f}_{x}(x,y)\) and \({f}_{y}(x,y),\) then use with \({x}_{0}=\pi \text{/}3\) and \({y}_{0}=\pi \text{/}4\text{:}\)

    \[\begin{array}{lll}{f}_{x}(x,y) & = & 2\ \text{cos}(2x)\text{cos}(3y) \\ {f}_{y}(x,y) & = & -3\ \text{sin}(2x)\text{sin}(3y) \\ f(\frac{\pi }{3},\frac{\pi }{4}) & = & \text{sin}(2(\frac{\pi }{3}))\text{cos}(3(\frac{\pi }{4}))=(\frac{\sqrt{3}}{2})(-\frac{\sqrt{2}}{2})=-\frac{\sqrt{6}}{4} \\ {f}_{x}(\frac{\pi }{3},\frac{\pi }{4}) & = & 2\ \text{cos}(2(\frac{\pi }{3}))\text{cos}(3(\frac{\pi }{4}))=2(-\frac{1}{2})(-\frac{\sqrt{2}}{2})=\frac{\sqrt{2}}{2} \\ {f}_{y}(\frac{\pi }{3},\frac{\pi }{4}) & = & -3\ \text{sin}(2(\frac{\pi }{3}))\text{sin}(3(\frac{\pi }{4}))=-3(\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2})=-\frac{3\sqrt{6}}{4}.\end{array}\]

    Then becomes

    \[\begin{array}{l} \\ z=f({x}_{0},{y}_{0})+{f}_{x}({x}_{0},{y}_{0})(x-{x}_{0})+{f}_{y}({x}_{0},{y}_{0})(y-{y}_{0}) \\ \\ z=-\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{2}(x-\frac{\pi }{3})-\frac{3\sqrt{6}}{4}(y-\frac{\pi }{4}) \\ z=\frac{\sqrt{2}}{2}x-\frac{3\sqrt{6}}{4}y-\frac{\sqrt{6}}{4}-\frac{\pi \sqrt{2}}{6}+\frac{3\pi \sqrt{6}}{16}.\end{array}\]

  4. Given the function \(f(x,y)=\sqrt{41-4{x}^{2}-{y}^{2}},\) approximate \(f(2.1,2.9)\) using point \((2,3)\) for \(({x}_{0},{y}_{0}).\) What is the approximate value of \(f(2.1,2.9)\) to four decimal places?

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    To apply , we first must calculate \(f({x}_{0},{y}_{0}),\) \({f}_{x}({x}_{0},{y}_{0}),\) and \({f}_{y}({x}_{0},{y}_{0})\) using \({x}_{0}=2\) and \({y}_{0}=3\text{:}\)

    \[\begin{array}{lll}f({x}_{0},{y}_{0}) & = & f(2,3)=\sqrt{41-4{(2)}^{2}-{(3)}^{2}}=\sqrt{41-16-9}=\sqrt{16}=4 \\ {f}_{x}(x,y) & = & -\frac{4x}{\sqrt{41-4{x}^{2}-{y}^{2}}}\ \text{so}\ {f}_{x}({x}_{0},{y}_{0})=-\frac{4(2)}{\sqrt{41-4{(2)}^{2}-{(3)}^{2}}}=-2 \\ {f}_{y}(x,y) & = & -\frac{y}{\sqrt{41-4{x}^{2}-{y}^{2}}}\ \text{so}\ {f}_{y}({x}_{0},{y}_{0})=-\frac{3}{\sqrt{41-4{(2)}^{2}-{(3)}^{2}}}=-\frac{3}{4}.\end{array}\]

    Now we substitute these values into :

    \[\begin{array}{ll}L(x,y) & =f({x}_{0},{y}_{0})+{f}_{x}({x}_{0},{y}_{0})(x-{x}_{0})+{f}_{y}({x}_{0},{y}_{0})(y-{y}_{0}) \\ & =4-2(x-2)-\frac{3}{4}(y-3) \\ & =\frac{41}{4}-2x-\frac{3}{4}y.\end{array}\]

    Last, we substitute \(x=2.1\) and \(y=2.9\) into \(L(x,y)\text{:}\)

    \[L(2.1,2.9)=\frac{41}{4}-2(2.1)-\frac{3}{4}(2.9)=10.25-4.2-2.175=3.875.\]

    The approximate value of \(f(2.1,2.9)\) to four decimal places is

    \[f(2.1,2.9)=\sqrt{41-4{(2.1)}^{2}-{(2.9)}^{2}}=\sqrt{14.95}\approx 3.8665,\]

    which corresponds to a \(0.2\text{\%}\) error in approximation.

  5. Given the function \(f(x,y)={e}^{5-2x+3y},\) approximate \(f(4.1,0.9)\) using point \((4,1)\) for \(({x}_{0},{y}_{0}).\) What is the approximate value of \(f(4.1,0.9)\) to four decimal places?

    답을 드러내세요

    \(L(x,y)=6-2x+3y,\) so \(L(4.1,0.9)=6-2(4.1)+3(0.9)=0.5\) \(f(4.1,0.9)={e}^{5-2(4.1)+3(0.9)}={e}^{-0.5}\approx 0.6065.\)

  6. Show that the function \(f(x,y)=2{x}^{2}-4y\) is differentiable at point \((2,-3).\)

    답을 드러내세요

    First, we calculate \(f({x}_{0},{y}_{0}),{f}_{x}({x}_{0},{y}_{0}),\ \text{and}\ {f}_{y}({x}_{0},{y}_{0})\) using \({x}_{0}=2\) and \({y}_{0}=-3,\) then we use :

    \[\begin{array}{lll}f(2,-3) & = & 2{(2)}^{2}-4(-3)=8+12=20 \\ {f}_{x}(2,-3) & = & 4(2)=8 \\ {f}_{y}(2,-3) & = & -4.\end{array}\]

    Therefore \({m}_{1}=8\) and \({m}_{2}=-4,\) and becomes

    \[\begin{array}{lll}f(x,y) & = & f(2,-3)+{f}_{x}(2,-3)(x-2)+{f}_{y}(2,-3)(y+3)+E(x,y) \\ 2{x}^{2}-4y & = & 20+8(x-2)-4(y+3)+E(x,y) \\ 2{x}^{2}-4y & = & 20+8x-16-4y-12+E(x,y) \\ 2{x}^{2}-4y & = & 8x-4y-8+E(x,y) \\ E(x,y) & = & 2{x}^{2}-8x+8.\end{array}\]

    Next, we calculate \(\underset{(x,y)\to ({x}_{0},{y}_{0})}{\text{lim}}\frac{E(x,y)}{\sqrt{{(x-{x}_{0})}^{2}+{(y-{y}_{0})}^{2}}}\text{:}\)

    \[\begin{array}{ll}\underset{(x,y)\to ({x}_{0},{y}_{0})}{\text{lim}}\frac{E(x,y)}{\sqrt{{(x-{x}_{0})}^{2}+{(y-{y}_{0})}^{2}}} & =\underset{(x,y)\to (2,-3)}{\text{lim}}\frac{2{x}^{2}-8x+8}{\sqrt{{(x-2)}^{2}+{(y+3)}^{2}}} \\ & =\underset{(x,y)\to (2,-3)}{\text{lim}}\frac{2({x}^{2}-4x+4)}{\sqrt{{(x-2)}^{2}+{(y+3)}^{2}}} \\ & =\underset{(x,y)\to (2,-3)}{\text{lim}}\frac{2{(x-2)}^{2}}{\sqrt{{(x-2)}^{2}+{(y+3)}^{2}}} \\ & \le \underset{(x,y)\to (2,-3)}{\text{lim}}\frac{2({(x-2)}^{2}+{(y+3)}^{2})}{\sqrt{{(x-2)}^{2}+{(y+3)}^{2}}} \\ & =\underset{(x,y)\to (2,-3)}{\text{lim}}2\sqrt{{(x-2)}^{2}+{(y+3)}^{2}} \\ & =0.\end{array}\]

    Since \(E(x,y)\ge 0\) for any value of \(x\ \text{or}\ y,\) the original limit must be equal to zero. Therefore, \(f(x,y)=2{x}^{2}-4y\) is differentiable at point \((2,-3).\)

  7. Show that the function \(f(x,y)=3x-4{y}^{2}\) is differentiable at point \((-1,2).\)

    답을 드러내세요

    \(f(-1,2)=-19,\ {f}_{x}(-1,2)=3,{f}_{y}(-1,2)=-16,E(x,y)=-4{(y-2)}^{2}.\)

    \(\begin{array}{ll}\underset{(x,y)\to ({x}_{0},{y}_{0})}{\text{lim}}\frac{E(x,y)}{\sqrt{{(x-{x}_{0})}^{2}+{(y-{y}_{0})}^{2}}} & =\underset{(x,y)\to (-1,2)}{\text{lim}}\frac{-4{(y-2)}^{2}}{\sqrt{{(x+1)}^{2}+{(y-2)}^{2}}} \\ & \le \underset{(x,y)\to (-1,2)}{\text{lim}}\frac{-4({(x+1)}^{2}+{(y-2)}^{2})}{\sqrt{{(x+1)}^{2}+{(y-2)}^{2}}} \\ & =\underset{(x,y)\to (-1,2)}{\text{lim}}-4\sqrt{{(x+1)}^{2}+{(y-2)}^{2}} \\ & =0.\end{array}\)

  8. Find the differential \(dz\) of the function \(f(x,y)=3{x}^{2}-2xy+{y}^{2}\) and use it to approximate \(\Delta z\) at point \((2,-3).\) Use \(\Delta x=0.1\) and \(\Delta y=-0.05.\) What is the exact value of \(\Delta z?\)

    답을 드러내세요

    First, we must calculate \(f({x}_{0},{y}_{0}),{f}_{x}({x}_{0},{y}_{0}),\ \text{and}\ {f}_{y}({x}_{0},{y}_{0})\) using \({x}_{0}=2\) and \({y}_{0}=-3\text{:}\)

    \[\begin{array}{lll}f({x}_{0},{y}_{0}) & = & f(2,-3)=3{(2)}^{2}-2(2)(-3)+{(-3)}^{2}=12+12+9=33 \\ {f}_{x}(x,y) & = & 6x-2y \\ {f}_{y}(x,y) & = & -2x+2y \\ {f}_{x}({x}_{0},{y}_{0}) & = & {f}_{x}(2,-3)=6(2)-2(-3)=12+6=18 \\ {f}_{y}({x}_{0},{y}_{0}) & = & {f}_{y}(2,-3)=-2(2)+2(-3)=-4-6=-10.\end{array}\]

    Then, we substitute these quantities into :

    \[\begin{array}{l}dz={f}_{x}({x}_{0},{y}_{0})dx+{f}_{y}({x}_{0},{y}_{0})dy \\ dz=18(0.1)-10(-0.05)=1.8+0.5=2.3.\end{array}\]

    This is the approximation to \(\Delta z=f({x}_{0}+\Delta x,{y}_{0}+\Delta y)-f({x}_{0},{y}_{0}).\) The exact value of \(\Delta z\) is given by

    \[\begin{array}{ll}\Delta z & =f({x}_{0}+\Delta x,{y}_{0}+\Delta y)-f({x}_{0},{y}_{0}) \\ & =f(2+0.1,-3-0.05)-f(2,-3) \\ & =f(2.1,-3.05)-f(2,-3) \\ & =2.3425.\end{array}\]

  9. Find the differential \(dz\) of the function \(f(x,y)=4{y}^{2}+{x}^{2}y-2xy\) and use it to approximate \(\Delta z\) at point \((1,-1).\) Use \(\Delta x=0.03\) and \(\Delta y=-0.02.\) What is the exact value of \(\Delta z?\)

    답을 드러내세요

    \(\begin{array}{lll}dz & = & 0.18 \\ \Delta z & = & f(1.03,-1.02)-f(1,-1)=0.180682\end{array}\)

  10. \(z=f(x,y)={x}^{3},(2,-1,8)\)

    답을 드러내세요

    \((\frac{\sqrt{145}}{145})(12i-k)\)

  11. \(\text{ln}(\frac{x}{y-z})=0\) when \(x=y=1\)

  12. \({x}^{2}+xy+{y}^{2}=3,P(-1,-1)\)

    답을 드러내세요

    Normal vector: \(i+j,\) tangent vector: \(i-j\)

  13. \({({x}^{2}+{y}^{2})}^{2}=9({x}^{2}-{y}^{2}),P(\sqrt{2},1)\)

  14. \(x{y}^{2}-2{x}^{2}+y+5x=6,P(4,2)\)

    답을 드러내세요

    Normal vector: \(7i-17j,\) tangent vector: \(17i+7j\)

  15. \(2{x}^{3}-{x}^{2}{y}^{2}=3x-y-7,P(1,-2)\)

  16. \(z{e}^{{x}^{2}-{y}^{2}}-3=0,\) \(P(2,2,3)\)

    답을 드러내세요

    \(\begin{array}{ll}\text{Normal vector} & -12i+12j-k \\ \text{Tangent vector} & 0i+j+12k\text{ or }0i+j-12k\end{array}\)

  17. \(-8x-3y-7z=-19,P(1,-1,2)\)

  18. \(z=-9{x}^{2}-3{y}^{2},P(2,1,-39)\)

    답을 드러내세요

    \(-36x-6y-z=-39\)

  19. \({x}^{2}+10xyz+{y}^{2}+8{z}^{2}=0,P(-1,-1,-1)\)

  20. \(z=\text{ln}(10{x}^{2}+2{y}^{2}+1),P(0,0,0)\)

    답을 드러내세요

    \(z=0\)

  21. \(z={e}^{7{x}^{2}+4{y}^{2}},\) \(P(0,0,1)\)

  22. \(xy+yz+zx=11,P(1,2,3)\)

    답을 드러내세요

    \(5x+4y+3z-22=0\)

  23. \({x}^{2}+4{y}^{2}={z}^{2},P(3,2,5)\)

  24. \({x}^{3}+{y}^{3}=3xyz,P(1,2,\frac{3}{2})\)

    답을 드러내세요

    \(4x-5y+4z=0\)

  25. \(z=axy,P(1,\frac{1}{a},1)\)

  26. \(z=\text{sin}\ x+\text{sin}\ y+\text{sin}(x+y),P(0,0,0)\)

    답을 드러내세요

    \(2x+2y-z=0\)

  27. \(z=h(x,y)=\text{ln}\sqrt{{x}^{2}+{y}^{2}},P(3,4)\)

  28. \(z={x}^{2}-2xy+{y}^{2},P(1,2,1)\)

    답을 드러내세요

    \(-2(x-1)+2(y-2)-(z-1)=0\)

  29. \(-3x+9y+4z=-4,P(1,-1,2)\)

  30. \(z=5{x}^{2}-2{y}^{2},P(2,1,18)\)

    답을 드러내세요

    \(x=20t+2,y=-4t+1,z=\text{-}t+18\)

  31. \({x}^{2}-8xyz+{y}^{2}+6{z}^{2}=0,P(1,1,1)\)

  32. \(z=\text{ln}(3{x}^{2}+7{y}^{2}+1),P(0,0,0)\)

    답을 드러내세요

    \(x=0,y=0,z=t\)

  33. \(z={e}^{4{x}^{2}+6{y}^{2}},P(0,0,1)\)

  34. \(z={x}^{2}-2xy+{y}^{2}\) at point \(P(1,2,1)\)

    답을 드러내세요

    \(x-1=2t;\ y-2=-2t;\ z-1=t\)

  35. The length of line segment \(AC\) is equal to what mathematical expression?

  36. The length of line segment \(BC\) is equal to what mathematical expression?

    답을 드러내세요

    The differential of the function \(z(x,y)=dz={f}_{x}dx+{f}_{y}dy\)

  37. Using the figure, explain what the length of line segment \(AB\) represents.

  38. Show that \(f(x,y)={e}^{xy}x\) is differentiable at point \((1,0).\)

    답을 드러내세요

    Using the definition of differentiability, we have \({e}^{xy}x\approx x+y.\)

  39. Find the total differential of the function \(w={e}^{y}\text{cos}(x)+{z}^{2}.\)

  40. Show that \(f(x,y)={x}^{2}+3y\) is differentiable at every point. In other words, show that \(\Delta z=f(x+\Delta x,y+\Delta y)-f(x,y)={f}_{x}\Delta x+{f}_{y}\Delta y+{\epsilon }_{1}\Delta x+{\epsilon }_{2}\Delta y,\) where both \({\epsilon }_{1}\) and \({\epsilon }_{2}\) approach zero as \((\Delta x,\Delta y)\) approaches \((0,0).\)

    답을 드러내세요

    \(\Delta z=2x\Delta x+3\Delta y+{(\Delta x)}^{2}.\) \({(\Delta x)}^{2}\to 0\) for small \(\Delta x\) and \(z\) satisfies the definition of differentiability.

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어떻게: Tangent Planes and Linear Approximations

  1. Determine the equation of a plane tangent to a given surface at a point.
  2. Use the tangent plane to approximate a function of two variables at a point.
  3. Explain when a function of two variables is differentiable.
  4. Use the total differential to approximate the change in a function of two variables.
  5. The analog of a tangent line to a curve is a tangent plane to a surface for functions of two variables.
  6. Tangent planes can be used to approximate values of functions near known values.
  7. A function is differentiable at a point if it is ”smooth” at that point (i.e., no corners or discontinuities exist at that point).
  8. The total differential can be used to approximate the change in a function

사람들이 묻는 질문

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.

이 페이지의 일부는 다음에서 변경되었습니다. OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). 여기서 압축하고 다시 설명; 오류는 우리의. 소스가 비상업적 사용을 위해 라이선스되어 있기 때문에, 이 레슨은 무료이며, 아무것도 판매하지 않으며, 동일한 라이선스로 공유됩니다.

에 더 Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems