Calculus: Related Rates Worksheet

Question 1

A spherical balloon is being inflated so that its volume increases at a rate of 120 cubic centimeters per second. How fast is the radius increasing when the radius is 10 centimeters?

Accepted Answer

The volume is V = (4/3)pi r^3, so dV/dt = 4pi r^2 dr/dt. Substituting dV/dt = 120 and r = 10 gives 120 = 400pi dr/dt, so dr/dt = 3/(10pi) centimeters per second.

Question 2

A ladder 13 feet long rests against a vertical wall. The bottom of the ladder slides away from the wall at 2 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 5 feet from the wall?

Accepted Answer

Let x be the distance from the wall and y be the height on the wall. Since x^2 + y^2 = 13^2, differentiating gives 2x dx/dt + 2y dy/dt = 0. When x = 5, y = 12, so 2(5)(2) + 2(12)dy/dt = 0, giving dy/dt = -5/6 feet per second. The top is sliding down at 5/6 feet per second.

Question 3

A conical tank has height 12 meters and radius 4 meters at the top. Water is being pumped into the tank at 6 cubic meters per minute. How fast is the water level rising when the water is 3 meters deep?

Accepted Answer

By similar triangles, r/h = 4/12 = 1/3, so r = h/3. The water volume is V = (1/3)pi r^2 h = (1/3)pi(h/3)^2h = pi h^3/27. Differentiating gives dV/dt = (pi h^2/9)dh/dt. Substituting dV/dt = 6 and h = 3 gives 6 = pi dh/dt, so dh/dt = 6/pi meters per minute.

Question 4

A square metal plate is heated so that each side length increases at 0.08 centimeters per second. How fast is the area increasing when each side is 15 centimeters long?

Accepted Answer

The area is A = s^2, so dA/dt = 2s ds/dt. Substituting s = 15 and ds/dt = 0.08 gives dA/dt = 2(15)(0.08) = 2.4 square centimeters per second.

Question 5

A circular oil spill expands so that its radius increases at 0.5 meters per minute. How fast is the area of the spill increasing when the radius is 20 meters?

Accepted Answer

The area is A = pi r^2, so dA/dt = 2pi r dr/dt. Substituting r = 20 and dr/dt = 0.5 gives dA/dt = 20pi square meters per minute.

Question 6

A 10-foot ladder leans against a wall. The top slides down at 1.5 feet per second. How fast is the bottom moving away from the wall when the top is 6 feet above the ground?

Accepted Answer

Let x be the bottom distance and y be the height. The equation is x^2 + y^2 = 100. Differentiating gives 2x dx/dt + 2y dy/dt = 0. When y = 6, x = 8, and dy/dt = -1.5, so 16 dx/dt + 12(-1.5) = 0. Therefore dx/dt = 18/16 = 9/8 feet per second.

Question 7

A plane flying horizontally at an altitude of 3 miles passes directly over a radar station. The plane moves away from the station at 480 miles per hour. How fast is the distance from the plane to the radar station increasing when the plane is 4 miles horizontally from the station?

Accepted Answer

Let x be the horizontal distance and s be the distance from the radar station to the plane. Then s^2 = x^2 + 3^2. Differentiating gives 2s ds/dt = 2x dx/dt. When x = 4, s = 5, and dx/dt = 480, so ds/dt = (4)(480)/5 = 384 miles per hour.

Question 8

A rectangle has length increasing at 3 centimeters per second and width decreasing at 2 centimeters per second. How fast is the area changing when the length is 20 centimeters and the width is 8 centimeters?

Accepted Answer

The area is A = LW. Differentiating gives dA/dt = L dW/dt + W dL/dt. Substituting L = 20, W = 8, dL/dt = 3, and dW/dt = -2 gives dA/dt = 20(-2) + 8(3) = -16 square centimeters per second. The area is decreasing at 16 square centimeters per second.

Question 9

A cube is expanding so that its volume increases at 54 cubic inches per second. How fast is the edge length increasing when each edge is 3 inches long?

Accepted Answer

The volume is V = s^3, so dV/dt = 3s^2 ds/dt. Substituting dV/dt = 54 and s = 3 gives 54 = 27 ds/dt, so ds/dt = 2 inches per second.

Question 10

A streetlight is 18 feet tall. A 6-foot-tall person walks away from the streetlight at 4 feet per second. How fast is the tip of the person's shadow moving away from the streetlight?

Accepted Answer

Let x be the person's distance from the light and y be the shadow length. The tip of the shadow is x + y from the light. Similar triangles give 18/(x + y) = 6/y, so 18y = 6x + 6y and 12y = 6x, meaning y = x/2. Then dy/dt = 2 feet per second. The tip moves at d(x + y)/dt = dx/dt + dy/dt = 4 + 2 = 6 feet per second.

Question 11

Two cars leave the same intersection at the same time. One travels north at 60 miles per hour, and the other travels east at 80 miles per hour. How fast is the distance between them increasing after 2 hours?

Accepted Answer

Let x = 80t, y = 60t, and s be the distance between the cars. Since s^2 = x^2 + y^2, at t = 2 we have x = 160, y = 120, and s = 200. Differentiating gives 2s ds/dt = 2x dx/dt + 2y dy/dt. Substituting gives 200 ds/dt = 160(80) + 120(60), so ds/dt = 100 miles per hour.

Question 12

A cylindrical tank has radius 5 meters. Water drains from the tank at 10 cubic meters per minute. How fast is the water height changing?

Accepted Answer

The volume is V = pi r^2 h. Since r = 5 is constant, V = 25pi h. Differentiating gives dV/dt = 25pi dh/dt. Because the tank is draining, dV/dt = -10, so dh/dt = -10/(25pi) = -2/(5pi) meters per minute. The water height is decreasing at 2/(5pi) meters per minute.

Question 13

A particle moves along the curve y = x^2 + 1. At the moment when x = 3, the x-coordinate is increasing at 2 units per second. How fast is the y-coordinate changing at that moment?

Accepted Answer

Differentiate y = x^2 + 1 with respect to time to get dy/dt = 2x dx/dt. Substituting x = 3 and dx/dt = 2 gives dy/dt = 12 units per second.

Question 14

Sand falls onto a conical pile at a rate of 20 cubic feet per minute. The pile's height is always twice its radius. How fast is the height increasing when the pile is 6 feet high?

Accepted Answer

The cone volume is V = (1/3)pi r^2 h. Since h = 2r, r = h/2. Thus V = (1/3)pi(h/2)^2h = pi h^3/12. Differentiating gives dV/dt = (pi h^2/4)dh/dt. Substituting dV/dt = 20 and h = 6 gives 20 = 9pi dh/dt, so dh/dt = 20/(9pi) feet per minute.

Question 15

A camera is located 100 meters from a straight road. A car passes the point on the road closest to the camera and travels away at 25 meters per second. How fast must the camera's viewing angle turn when the car is 100 meters from that closest point on the road?

Accepted Answer

Let x be the car's distance along the road from the closest point and theta be the camera angle from the perpendicular line to the road. Then tan(theta) = x/100. Differentiating gives sec^2(theta) dtheta/dt = (1/100) dx/dt. When x = 100, tan(theta) = 1, so theta = pi/4 and sec^2(theta) = 2. Thus 2 dtheta/dt = 25/100, so dtheta/dt = 1/8 radians per second.

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