Related rates

Exercices (40)

Essayez chacun d'abord sur papier. Révélez la réponse pour vérifier; les vérifiés peuvent être ouverts dans le solveur pour chaque étape.

1. A cylindrical water tank on top of a city water tower has a radius of 8 meters and a height of 15 meters. The amount of water in the tank changes as water is being pumped in or out, depending on the time of day and the local water usage; it is usually completely refilled by about 2am, and then starts to drain around 7am as people wake up and do morning routines (e.g., showering, cooking, etc.).

   1. What is the maximum amount of water the tank can hold? Note: the formula for the volume of a cylinder is \(V = \pi r^2 h\).

   2. Draw what you imagine the water in the tanks would look like at 2am, at 8am, and at 11am.

   3. Which measurements of the water are different among your different pictures? Which measurements are the same?

   4. Determine a formula for the volume of water in the tank as a function of only one variable. What is the varying quantity on which your volume formula depends?

   5. Draw a graph relating the volume of water to the water level, and a second graph relating the volume of water to time.

   Carefully organize your responses on separate paper, labeling each part, and upload your file at the end of the response form.

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2. A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow. If the rope is being pulled in at a rate of 2 feet per second, how fast is the boat approaching the dock when the length of rope from bow to pulley is 13 feet?

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   Révèle la réponse

   Using the given information, we construct the figure shown below.

   As pictured, we know that \(5\) is the vertical height from the pulley to the level of the bow of the boat, and we let \(z\) be the length of the rope from the pulley to the bow of the boat, and \(x\) the horizontal distance from the dock to the bow of the boat.

   We are given that the rope is being pulled in at \(2\) feet per second, and thus \(\frac{dz}{dt} = -2\) feet per second. Because we want to know how fast the boat is approaching the dock when the length of rope from bow to pulley is 13 feet, we want to know \(\left. \frac{dx}{dt} \right|_{z=13}\). Thus, we need to relate the changing quantities \(z\) and \(x\).

   Because the rope, the post (vertically extended), and the horizontal distance from the bow of the boat to the post on the dock form a right triangle at all times, it follows that \[\begin{aligned}\end{aligned}\]. Having now related \(z\) and \(x\), we differentiate this equation with respect to \(t\). By the chain rule, we now see that \[\begin{aligned}\end{aligned}\]. At the instant \(z = 13\), \(x^2 + 5^2 = 13^2\), and thus \(x = 12\). Using all of the given information at the instant \(z = 13\) (including that \(\frac{dz}{dt} = -2\)), \[\begin{aligned}\end{aligned}\]. Solving for \(\left. \frac{dx}{dt} \right|_{z = 13}\), \[\begin{aligned}\end{aligned}\] feet per second. Thus the boat is approaching the dock at a rate of \(\frac{13}{6} \approx 2.167\) feet per second.

3. A swimming pool is \(60\) feet long and \(25\) feet wide. Its depth varies uniformly from \(3\) feet at the shallow end to \(15\) feet at the deep end, as shown in the Figure.

   Suppose the pool has been emptied and is now being filled with water at a rate of \(800\) cubic feet per minute. At what rate is the depth of water (measured at the deepest point of the pool) increasing when it is \(5\) feet deep at that end? Over time, describe how the depth of the water will increase: at an increasing rate, at a decreasing rate, or at a constant rate. Explain.

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   Révèle la réponse

   The variables in this problem are the volume \(V\) of water in the pool and the depth \(y\) of water in the pool (measured at the deepest point of the pool); each is implicitly a function of time, \(t\). We are given that \(\frac{dV}{dt}\) is a constant \(800\) cubic feet per minute and want to find \(\frac{dh}{dt}\bigm|_{h=5}\). To do this, we first need to relate \(V\) and \(h\). The volume of the water in the pool at height \(h \lt 12\) is the volume of the triangular cross sectional area of height \(h\) (as shown in the figure below) times the width (25 feet) of the pool. The height of the triangular cross section is \(h\) and the length is distance of the dotted line indicated in the figure.

   Placing the cross section on a coordinate system as in the figure, the hypotenuse of the triangle is the line connecting the points \((0,-15)\) and \((60, -3)\). This line has slope \(\frac{12}{60} = 0.2\) and \(y\)-intercept \((0,-15)\). So the equation of this line is \(y = 0.2x-15\). It is the \(x\) coordinate of the point on this line corresponding to the \(y\)-coordinate \(h-15\) that is the length of the triangle whose height is \(h\). So the length \(l\) is \(l = \frac{(h-15)+15}{0.2} = 5h\). Thus, the volume of water in the pool at height \(h\) is \[\begin{aligned}\end{aligned}\]. Both \(V\) and \(h\) are functions of time. Differentiating both sides of the equation with respect to \(t\) gives \[\begin{aligned}\end{aligned}\], so \[\begin{aligned}\end{aligned}\]. When the depth of water in the pool is \(5\) feet, then the depth is increasing at the rate \[\begin{aligned}\end{aligned}\] feet per minute. Since \(\frac{dh}{dt}\) is inversely proportional to \(h\), the rate at which the depth of the water increases slows as \(h\) increases, so the depth of the water is increasing at a decreasing rate.

4. A baseball diamond is a square with sides \(90\) feet long. Suppose a baseball player is advancing from second to third base at the rate of \(24\) feet per second, and an umpire is standing on home plate. Let \(\theta\) be the angle between the third baseline and the line of sight from the umpire to the runner. How fast is \(\theta\) changing when the runner is \(30\) feet from third base?

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   Révèle la réponse

   Let \(x\) represent the distance from the runner to third base along the path from second to third, let \(z\) be the distance from the umpire at home plate to the runner, and call \(\theta\) the angle between the third baseline and the line of sight from the umpire to the runner. See the figure below.

   Each of \(x\), \(z\), and \(\theta\) is implicitly a function of \(t\). We are given that the runner is advancing at a rate of \(24\) feet per second, so \(\frac{dx}{dt} = -24\). Since we want to know \(\left. \frac{d\theta}{dt} \right|_{x = 30}\), we want to relate \(x\) and \(\theta\).

   We first observe that by definition, \[\begin{aligned}\end{aligned}\]. Having related \(x\) and \(\theta\), we can differentiate both sides of the preceding equation with respect to \(t\). By the chain rule, \[\begin{aligned}\end{aligned}\], and thus \[\begin{aligned}\end{aligned}\] At the instant \(x = 30\), since \(x^2 + 90^2 = z^2\), \(z = \sqrt{90^2 + 30^2} = \sqrt{9000} = 30\sqrt{10}\). Hence, at this same instance, \(\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{30\sqrt{10}}{90} = \frac{\sqrt{10}}{3}\). Recall we also know that \(\frac{dx}{dt} = -24\). Applying this information to the equation that relates the rates of change of \(x\) and \(\theta\), \[\begin{aligned}\end{aligned}\] radians per second.

5. Sand is being dumped off a conveyor belt onto a pile in such a way that the pile forms in the shape of a cone whose radius is always equal to its height. Assuming that the sand is being dumped at a rate of \(10\) cubic feet per minute, how fast is the height of the pile changing when there are \(1000\) cubic feet on the pile?

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   We can draw a figure similar to Figure and let \(h\) be the height of the pile at time \(t\) and \(r\) the pile's base radius. Given that \(\frac{dV}{dt} = 10\) cubic feet per minute, we want to know \(\left. \frac{dh}{dt} \right|_{V = 1000}\).

   We first relate \(V\) and \(h\). We know that \(V = \frac{1}{3} \pi r^2 h\), as well as that for every time, \(t\), the height and radius are equal so \(r = h\). Thus, \[\begin{aligned}\end{aligned}\] Differentiating both sides of this equation with respect to \(t\) \[\begin{aligned}\end{aligned}\]. We note that at the instant \(V = 1000\), \(1000 = \frac{1}{3} \pi h^3\), so \(h^3 = \frac{3000}{\pi}\) and therefore \(h = \sqrt[3]{\frac{3000}{\pi}}\). Recall also that \(\frac{dV}{dt} = 10\). Applying all of this information at the instant \(V = 1000\) in the equation that relates \(\frac{dV}{dt}\) and \(\frac{dh}{dt}\), we have \[\begin{aligned}\end{aligned}\]. Thus, \[\begin{aligned}\end{aligned}\] feet per minute.

6. A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow. If the rope is being pulled in at a rate of 2 feet per second, how fast is the boat approaching the dock when the length of rope from bow to pulley is 13 feet?

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   Révèle la réponse

   Using the given information, we construct the figure shown below.

   As pictured, we know that \(5\) is the vertical height from the pulley to the level of the bow of the boat, and we let \(z\) be the length of the rope from the pulley to the bow of the boat, and \(x\) the horizontal distance from the dock to the bow of the boat.

   We are given that the rope is being pulled in at \(2\) feet per second, and thus \(\frac{dz}{dt} = -2\) feet per second. Because we want to know how fast the boat is approaching the dock when the length of rope from bow to pulley is 13 feet, we want to know \(\left. \frac{dx}{dt} \right|_{z=13}\). Thus, we need to relate the changing quantities \(z\) and \(x\).

   Because the rope, the post (vertically extended), and the horizontal distance from the bow of the boat to the post on the dock form a right triangle at all times, it follows that \[\begin{aligned}\end{aligned}\]. Having now related \(z\) and \(x\), we differentiate this equation with respect to \(t\). By the chain rule, we now see that \[\begin{aligned}\end{aligned}\]. At the instant \(z = 13\), \(x^2 + 5^2 = 13^2\), and thus \(x = 12\). Using all of the given information at the instant \(z = 13\) (including that \(\frac{dz}{dt} = -2\)), \[\begin{aligned}\end{aligned}\]. Solving for \(\left. \frac{dx}{dt} \right|_{z = 13}\), \[\begin{aligned}\end{aligned}\] feet per second. Thus the boat is approaching the dock at a rate of \(\frac{13}{6} \approx 2.167\) feet per second.

7. A swimming pool is \(60\) feet long and \(25\) feet wide. Its depth varies uniformly from \(3\) feet at the shallow end to \(15\) feet at the deep end, as shown in the Figure.

   Suppose the pool has been emptied and is now being filled with water at a rate of \(800\) cubic feet per minute. At what rate is the depth of water (measured at the deepest point of the pool) increasing when it is \(5\) feet deep at that end? Over time, describe how the depth of the water will increase: at an increasing rate, at a decreasing rate, or at a constant rate. Explain.

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   Révèle la réponse

   The variables in this problem are the volume \(V\) of water in the pool and the depth \(y\) of water in the pool (measured at the deepest point of the pool); each is implicitly a function of time, \(t\). We are given that \(\frac{dV}{dt}\) is a constant \(800\) cubic feet per minute and want to find \(\frac{dh}{dt}\bigm|_{h=5}\). To do this, we first need to relate \(V\) and \(h\). The volume of the water in the pool at height \(h \lt 12\) is the volume of the triangular cross sectional area of height \(h\) (as shown in the figure below) times the width (25 feet) of the pool. The height of the triangular cross section is \(h\) and the length is distance of the dotted line indicated in the figure.

   Placing the cross section on a coordinate system as in the figure, the hypotenuse of the triangle is the line connecting the points \((0,-15)\) and \((60, -3)\). This line has slope \(\frac{12}{60} = 0.2\) and \(y\)-intercept \((0,-15)\). So the equation of this line is \(y = 0.2x-15\). It is the \(x\) coordinate of the point on this line corresponding to the \(y\)-coordinate \(h-15\) that is the length of the triangle whose height is \(h\). So the length \(l\) is \(l = \frac{(h-15)+15}{0.2} = 5h\). Thus, the volume of water in the pool at height \(h\) is \[\begin{aligned}\end{aligned}\]. Both \(V\) and \(h\) are functions of time. Differentiating both sides of the equation with respect to \(t\) gives \[\begin{aligned}\end{aligned}\], so \[\begin{aligned}\end{aligned}\]. When the depth of water in the pool is \(5\) feet, then the depth is increasing at the rate \[\begin{aligned}\end{aligned}\] feet per minute. Since \(\frac{dh}{dt}\) is inversely proportional to \(h\), the rate at which the depth of the water increases slows as \(h\) increases, so the depth of the water is increasing at a decreasing rate.

8. A baseball diamond is a square with sides \(90\) feet long. Suppose a baseball player is advancing from second to third base at the rate of \(24\) feet per second, and an umpire is standing on home plate. Let \(\theta\) be the angle between the third baseline and the line of sight from the umpire to the runner. How fast is \(\theta\) changing when the runner is \(30\) feet from third base?

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   Révèle la réponse

   Let \(x\) represent the distance from the runner to third base along the path from second to third, let \(z\) be the distance from the umpire at home plate to the runner, and call \(\theta\) the angle between the third baseline and the line of sight from the umpire to the runner. See the figure below.

   Each of \(x\), \(z\), and \(\theta\) is implicitly a function of \(t\). We are given that the runner is advancing at a rate of \(24\) feet per second, so \(\frac{dx}{dt} = -24\). Since we want to know \(\left. \frac{d\theta}{dt} \right|_{x = 30}\), we want to relate \(x\) and \(\theta\).

   We first observe that by definition, \[\begin{aligned}\end{aligned}\]. Having related \(x\) and \(\theta\), we can differentiate both sides of the preceding equation with respect to \(t\). By the chain rule, \[\begin{aligned}\end{aligned}\], and thus \[\begin{aligned}\end{aligned}\] At the instant \(x = 30\), since \(x^2 + 90^2 = z^2\), \(z = \sqrt{90^2 + 30^2} = \sqrt{9000} = 30\sqrt{10}\). Hence, at this same instance, \(\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{30\sqrt{10}}{90} = \frac{\sqrt{10}}{3}\). Recall we also know that \(\frac{dx}{dt} = -24\). Applying this information to the equation that relates the rates of change of \(x\) and \(\theta\), \[\begin{aligned}\end{aligned}\] radians per second.

9. Sand is being dumped off a conveyor belt onto a pile in such a way that the pile forms in the shape of a cone whose radius is always equal to its height. Assuming that the sand is being dumped at a rate of \(10\) cubic feet per minute, how fast is the height of the pile changing when there are \(1000\) cubic feet on the pile?

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   Révèle la réponse

   We can draw a figure similar to Figure and let \(h\) be the height of the pile at time \(t\) and \(r\) the pile's base radius. Given that \(\frac{dV}{dt} = 10\) cubic feet per minute, we want to know \(\left. \frac{dh}{dt} \right|_{V = 1000}\).

   We first relate \(V\) and \(h\). We know that \(V = \frac{1}{3} \pi r^2 h\), as well as that for every time, \(t\), the height and radius are equal so \(r = h\). Thus, \[\begin{aligned}\end{aligned}\] Differentiating both sides of this equation with respect to \(t\) \[\begin{aligned}\end{aligned}\]. We note that at the instant \(V = 1000\), \(1000 = \frac{1}{3} \pi h^3\), so \(h^3 = \frac{3000}{\pi}\) and therefore \(h = \sqrt[3]{\frac{3000}{\pi}}\). Recall also that \(\frac{dV}{dt} = 10\). Applying all of this information at the instant \(V = 1000\) in the equation that relates \(\frac{dV}{dt}\) and \(\frac{dh}{dt}\), we have \[\begin{aligned}\end{aligned}\]. Thus, \[\begin{aligned}\end{aligned}\] feet per minute.

10. A spherical balloon is being filled with air at the constant rate of \(2{\ \text{cm}}^{3}\text{/}\text{sec}\) (). How fast is the radius increasing when the radius is \(3\ \text{cm}?\)

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    Révèle la réponse

    The volume of a sphere of radius \(r\) centimeters is

    \[V=\frac{4}{3}\pi {r}^{3}{\text{cm}}^{3}.\]

    Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, \(t\) seconds after beginning to fill the balloon with air, the volume of air in the balloon is

    \[V(t)=\frac{4}{3}\pi {[r(t)]}^{3}{\text{cm}}^{3}.\]

    Differentiating both sides of this equation with respect to time and applying the chain rule, we see that the rate of change in the volume is related to the rate of change in the radius by the equation

    \[V'(t)=4\pi {[r(t)]}^{2}{r}^{'}(t).\]

    The balloon is being filled with air at the constant rate of 2 cm³/sec, so \(V'(t)=2{\ \text{cm}}^{3}\text{/}\text{sec}.\) Therefore,

    \[2{\text{cm}}^{3}\text{/}\text{sec}=(4\pi {[r(t)]}^{2}{\text{cm}}^{2})\cdot (r'(t)\text{cm/s})\text{,}\]

    which implies

    \[r'(t)=\frac{1}{2\pi {[r(t)]}^{2}}\ \text{cm/sec}.\]

    When the radius \(r=3\ \text{cm,}\)

    \[r'(t)=\frac{1}{18\pi }\ \text{cm/sec}.\]

11. What is the instantaneous rate of change of the radius when \(r=6\ \text{cm}?\)

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    Révèle la réponse

    \(\frac{1}{72\pi }\ \text{cm/sec},\) or approximately 0.0044 cm/sec

12. An airplane is flying overhead at a constant elevation of \(4000\ \text{ft}.\) A man is viewing the plane from a position \(3000\ \text{ft}\) from the base of a radio tower. The airplane is flying horizontally away from the man. If the plane is flying at the rate of \(600\ \text{ft/sec},\) at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower?

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    Révèle la réponse

    Step 1. Draw a picture, introducing variables to represent the different quantities involved.

    As shown, \(x\) denotes the distance between the man and the position on the ground directly below the airplane. The variable \(s\) denotes the distance between the man and the plane. Note that both \(x\) and \(s\) are functions of time. We do not introduce a variable for the height of the plane because it remains at a constant elevation of \(4000\ \text{ft}.\) Since an object’s height above the ground is measured as the shortest distance between the object and the ground, the line segment of length 4000 ft is perpendicular to the line segment of length \(x\) feet, creating a right triangle.

    Step 2. Since \(x\) denotes the horizontal distance between the man and the point on the ground below the plane, \(dx\text{/}dt\) represents the speed of the plane. We are told the speed of the plane is 600 ft/sec. Therefore, \(\frac{dx}{dt}=600\) ft/sec. Since we are asked to find the rate of change in the distance between the man and the plane when the plane is directly above the radio tower, we need to find \(ds\text{/}dt\) when \(x=3000\ \text{ft}.\)

    Step 3. From the figure, we can use the Pythagorean theorem to write an equation relating \(x\) and \(s\text{:}\)

    \[{[x(t)]}^{2}+{4000}^{2}={[s(t)]}^{2}.\]

    Step 4. Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation

    \[x\frac{dx}{dt}=s\frac{ds}{dt}.\]

    Step 5. Find the rate at which the distance between the man and the plane is increasing when the plane is directly over the radio tower. That is, find \(\frac{ds}{dt}\) when \(x=3000\ \text{ft}.\) Since the speed of the plane is \(600\ \text{ft/sec},\) we know that \(\frac{dx}{dt}=600\ \text{ft/sec}.\) We are not given an explicit value for \(s;\) however, since we are trying to find \(\frac{ds}{dt}\) when \(x=3000\ \text{ft},\) we can use the Pythagorean theorem to determine the distance \(s\) when \(x=3000\) and the height is \(4000\ \text{ft}.\) Solving the equation

    \[{3000}^{2}+{4000}^{2}={s}^{2}\]

    for \(s,\) we have \(s=5000\ \text{ft}\) at the time of interest. Using these values, we conclude that \(ds\text{/}dt\) is a solution of the equation

    \[(3000)(600)=(5000)\cdot \frac{ds}{dt}.\]

    Therefore,

    \[\frac{ds}{dt}=\frac{3000\cdot 600}{5000}=360\ \text{ft/sec}.\]

    Note: When solving related-rates problems, it is important not to substitute values for the variables too soon. For example, in step 3, we related the variable quantities \(x(t)\) and \(s(t)\) by the equation

    \[{[x(t)]}^{2}+{4000}^{2}={[s(t)]}^{2}.\]

    Since the plane remains at a constant height, it is not necessary to introduce a variable for the height, and we are allowed to use the constant 4000 to denote that quantity. However, the other two quantities are changing. If we mistakenly substituted \(x(t)=3000\) into the equation before differentiating, our equation would have been

    \[{3000}^{2}+{4000}^{2}={[s(t)]}^{2}.\]

    After differentiating, our equation would become

    \[0=s(t)\frac{ds}{dt}.\]

    As a result, we would incorrectly conclude that \(\frac{ds}{dt}=0.\)

13. What is the speed of the plane if the distance between the person and the plane is increasing at the rate of \(300\ \text{ft/sec}?\)

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    Révèle la réponse

    \(500\ \text{ft/sec}\)

14. A rocket is launched so that it rises vertically. A camera is positioned \(5000\ \text{ft}\) from the launch pad. When the rocket is \(1000\ \text{ft}\) above the launch pad, its velocity is \(600\ \text{ft/sec}.\) Find the necessary rate of change of the camera’s angle as a function of time so that it stays focused on the rocket.

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    Révèle la réponse

    Step 1. Draw a picture introducing the variables.

    Let \(h\) denote the height of the rocket above the launch pad and \(\theta\) be the angle between the camera lens and the ground.

    Step 2. We are trying to find the rate of change in the angle of the camera with respect to time when the rocket is 1000 ft off the ground. That is, we need to find \(\frac{d\theta }{dt}\) when \(h=1000\ \text{ft}.\) At that time, we know the velocity of the rocket is \(\frac{dh}{dt}=600\ \text{ft/sec}.\)

    Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: \(h\) and \(\theta .\) How can we create such an equation? Using the fact that we have drawn a right triangle, it is natural to think about trigonometric functions. Recall that \(\text{tan}\ \theta\) is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have

    \[\text{tan}\ \theta =\frac{h}{5000}.\]

    This gives us the equation

    \[h=5000\ \text{tan}\ \theta .\]

    Step 4. Differentiating this equation with respect to time \(t,\) we obtain

    \[\frac{dh}{dt}=5000\ {\text{sec}}^{2}\theta \ \frac{d\theta }{dt}.\]

    Step 5. We want to find \(\frac{d\theta }{dt}\) when \(h=1000\ \text{ft}.\) At this time, we know that \(\frac{dh}{dt}=600\ \text{ft/sec}.\) We need to determine \({\text{sec}}^{2}\theta .\) Recall that \(\text{sec}\ \theta\) is the ratio of the length of the hypotenuse to the length of the adjacent side. We know the length of the adjacent side is \(5000\ \text{ft}.\) To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is \(5000\ \text{ft},\) the length of the other leg is \(h=1000\ \text{ft},\) and the length of the hypotenuse is \(c\) feet as shown in the following figure.

    We see that

    \[{1000}^{2}+{5000}^{2}={c}^{2}\]

    and we conclude that the hypotenuse is

    \[c=1000\sqrt{26}\ \text{ft}.\]

    Therefore, when \(h=1000,\) we have

    \[{\text{sec}}^{2}\theta ={(\frac{1000\sqrt{26}}{5000})}^{2}=\frac{26}{25}.\]

    Recall from step 4 that the equation relating \(\frac{d\theta }{dt}\) to our known values is

    \[\frac{dh}{dt}=5000\ {\text{sec}}^{2}\theta \ \frac{d\theta }{dt}.\]

    When \(h=1000\ \text{ft},\) we know that \(\frac{dh}{dt}=600\ \text{ft/sec}\) and \({\text{sec}}^{2}\theta =\frac{26}{25}.\) Substituting these values into the previous equation, we arrive at the equation

    \[600=5000(\frac{26}{25})\frac{d\theta }{dt}\text{.}\]

    Therefore, \(\frac{d\theta }{dt}=\frac{3}{26}\ \text{rad/sec}.\)

15. What rate of change is necessary for the elevation angle of the camera if the camera is placed on the ground at a distance of \(4000\ \text{ft}\) from the launch pad and the velocity of the rocket is 500 ft/sec when the rocket is \(2000\ \text{ft}\) off the ground?

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    Révèle la réponse

    \(\frac{1}{10}\ \text{rad/sec}\)

16. Water is draining from the bottom of a cone-shaped funnel at the rate of \(0.0{3\ \text{ft}}^{3}\text{/sec}.\) The height of the funnel is 2 ft and the radius at the top of the funnel is \(1\ \text{ft}.\) At what rate is the height of the water in the funnel changing when the height of the water is \(\frac{1}{2}\ \text{ft}?\)

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    Révèle la réponse

    Step 1: Draw a picture introducing the variables.

    Let \(h\) denote the height of the water in the funnel, \(r\) denote the radius of the water at its surface, and \(V\) denote the volume of the water.

    Step 2: We need to determine \(\frac{dh}{dt}\) when \(h=\frac{1}{2}\ \text{ft}.\) We know that \(\frac{dV}{dt}=-0.03\ {\text{ft}}^{3}\text{/sec}.\)

    Step 3: The volume of water in the cone is

    \[V=\frac{1}{3}\pi {r}^{2}h.\]

    From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, \(\frac{r}{h}=\frac{1}{2}\) or \(r=\frac{h}{2}.\) Using this fact, the equation for volume can be simplified to

    \[V=\frac{1}{3}\pi {(\frac{h}{2})}^{2}h=\frac{\pi }{12}\ {h}^{3}.\]

    Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time \(t,\) we obtain

    \[\frac{dV}{dt}=\frac{\pi }{4}\ {h}^{2}\frac{dh}{dt}.\]

    Step 5: We want to find \(\frac{dh}{dt}\) when \(h=\frac{1}{2}\ \text{ft}.\) Since water is leaving at the rate of \(0.0{3\ \text{ft}}^{3}\text{/sec},\) we know that \(\frac{dV}{dt}=-0.03{\ \text{ft}}^{3}\text{/sec}.\) Therefore,

    \[-0.03=\frac{\pi }{4}{(\frac{1}{2})}^{2}\frac{dh}{dt}\ ,\]

    which implies

    \[-0.03=\frac{\pi }{16}\ \frac{dh}{dt}.\]

    It follows that

    \[\frac{dh}{dt}=-\frac{0.48}{\pi }=-0.153\ \text{ft/sec}.\]

17. At what rate is the height of the water changing when the height of the water is \(\frac{1}{4}\ \text{ft}?\)

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    Révèle la réponse

    \(-0.61\ \text{ft/sec}\)

18. Find \(\frac{dy}{dt}\) at \(x=1\) and \(y={x}^{2}+3\) if \(\frac{dx}{dt}=4.\)

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    \(8\)

19. Find \(\frac{dx}{dt}\) at \(x=-2\) and \(y=2{x}^{2}+1\) if \(\frac{dy}{dt}=-1.\)

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20. Find \(\frac{dz}{dt}\) at \((x,y)=(1,3)\) and \({z}^{2}={x}^{2}+{y}^{2}\) if \(\frac{dx}{dt}=4\) and \(\frac{dy}{dt}=3.\)

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    \(\pm \frac{13}{\sqrt{10}}\)

21. [T] If two electrical resistors are connected in parallel, the total resistance (measured in ohms, denoted by the Greek capital letter omega, \(\text{Ω})\) is given by the equation \(\frac{1}{R}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}.\) If \({R}_{1}\) is increasing at a rate of \(0.5\ \text{Ω}\text{/}\text{min}\) and \({R}_{2}\) decreases at a rate of \(1.1\text{Ω/min},\) at what rate does the total resistance change when \({R}_{1}=20\text{Ω}\) and \({R}_{2}=50\text{Ω}\)?

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22. A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall?

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    \(2\sqrt{3}\) ft/sec

23. A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially \(20\ \text{ft}\) away from the wall, how fast does the ladder move up the wall \(5\ \text{sec}\) after we start pushing?

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24. Two airplanes are flying in the air at the same height: airplane A is flying east at 250 mi/h and airplane B is flying north at \(300\ \text{mi/h}.\) If they are both heading to the same airport, located 30 miles east of airplane A and 40 miles north of airplane B, at what rate is the distance between the airplanes changing?

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    The distance is decreasing at \(390\ \text{mi/h}.\)

25. You and a friend are riding your bikes to a restaurant that you think is east; your friend thinks the restaurant is north. You both leave from the same point, with you riding at 16 mph east and your friend riding \(12\ \text{mph}\) north. After you traveled \(4\ \text{mi,}\) at what rate is the distance between you changing?

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26. Two buses are driving along parallel freeways that are \(5\ \text{mi}\) apart, one heading east and the other heading west. Assuming that each bus drives a constant \(55\ \text{mph,}\) find the rate at which the distance between the buses is changing when they are \(13\ \text{mi}\) apart, heading toward each other.

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    The distance between them shrinks at a rate of \(\frac{1320}{13}\approx 101.5\ \text{mph}.\)

27. A 6-ft-tall person walks away from a 10-ft lamppost at a constant rate of \(3\ \text{ft/sec}.\) What is the rate that the tip of the shadow moves away from the pole when the person is \(10\ \text{ft}\) away from the pole?

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28. Using the previous problem, what is the rate at which the tip of the shadow moves away from the person when the person is 10 ft from the pole?

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    \(\frac{9}{2}\) ft/sec

29. A 5-ft-tall person walks toward a wall at a rate of 2 ft/sec. A spotlight is located on the ground 40 ft from the wall. How fast does the height of the person’s shadow on the wall change when the person is 10 ft from the wall?

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30. Using the previous problem, what is the rate at which the shadow changes when the person is 10 ft from the wall, if the person is walking away from the wall at a rate of 2 ft/sec?

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    It grows at a rate \(\frac{4}{9}\) ft/sec

31. A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/sec. You are running on the ground starting directly under the helicopter at a rate of 10 ft/sec. Find the rate of change of the distance between the helicopter and yourself after 5 sec.

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32. Using the previous problem, what is the rate at which the distance between you and the helicopter is changing when the helicopter has risen to a height of 60 ft in the air, assuming that, initially, it was 30 ft above you?

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    The distance is increasing at \(\frac{(135\sqrt{26})}{26}\) ft/sec

33. The side of a cube increases at a rate of \(\frac{1}{2}\) m/sec. Find the rate at which the volume of the cube increases when the side of the cube is 4 m.

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34. The volume of a cube decreases at a rate of 10 m³/s. Find the rate at which the side of the cube changes when the side of the cube is 2 m.

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    \(-\frac{5}{6}\) m/sec

35. The radius of a circle increases at a rate of \(2\) m/sec. Find the rate at which the area of the circle increases when the radius is 5 m.

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36. The radius of a sphere decreases at a rate of \(3\) m/sec. Find the rate at which the surface area decreases when the radius is 10 m.

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    \(240\pi\) m²/sec

37. The radius of a sphere increases at a rate of \(1\) m/sec. Find the rate at which the volume increases when the radius is \(20\) m.

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38. The radius of a sphere is increasing at a rate of 9 cm/sec. Find the radius of the sphere when the volume and the radius of the sphere are increasing at the same numerical rate.

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    \(\frac{1}{2\sqrt{\pi }}\) cm

39. The base of a triangle is shrinking at a rate of 1 cm/min and the height of the triangle is increasing at a rate of 5 cm/min. Find the rate at which the area of the triangle changes when the height is 22 cm and the base is 10 cm.

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40. A triangle has two constant sides of length 3 ft and 5 ft. The angle between these two sides is increasing at a rate of 0.1 rad/sec. Find the rate at which the area of the triangle is changing when the angle between the two sides is \(\pi \text{/}6.\)

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    The area is increasing at a rate \(\frac{(3\sqrt{3})}{8}\ {\text{ft}}^{2}\text{/sec.}\)

Certaines parties de cette page sont adaptées de Boelkins, Active Calculus (CC BY-SA 4.0), OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), Wikipedia (CC BY-SA 4.0). Condensés et réexpliqués ici ; les erreurs sont les nôtres. Comme sa source est autorisée à des fins non commerciales, cette leçon est gratuite, ne vend rien et est partagée avec la même licence.

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LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests