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Solving Related Rates Problems in Calculus

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Ray is a Licensed Engineer in the Philippines. He loves to write any topic about mathematics and civil engineering.

What Are Related Rates?

Related Rates are Calculus problems that involve finding a rate at which a quantity changes by relating to other known values whose rates of change are known. For instance, if we pump air into a donut floater, both the radius and the balloon volume increase, and their growth rates are related. Both can be solved, but it is much easier to solve the volume rate of change. Take note that the related rates of change are generally expressed in terms of time.

How to Do Related Rates?

There are plenty of strategies on how to do related rates, but you must consider the necessary steps.

  1. Read and understand the problem carefully. According to the Principles of Problem Solving, the first step is always to understand the problem. It includes reading the related rates problem carefully, identifying the given, and identifying the unknown. If possible, try to read the problem at least two times to understand the situation entirely.
  2. Draw a diagram or sketch, if possible. Drawing a picture or representation of the given problem can help in visualizing and keeping everything organized.
  3. Introduce notations or symbols. Assign symbols or variables to all quantities that are functions of time.
  4. Express the given information and the necessary rate in terms of derivatives. Remember that rates of change are derivatives. Restate the given and the unknown as derivatives.
  5. Write an equation that relates the several quantities of the problem. Write an equation relating the quantities whose rates of change are known to the value whose rate of change is to be solved. It would help to the thought of a plan for connecting the given and the unknown. If necessary, use the geometry of the situation to eliminate one of the variables by substitution method.
  6. Use the chain rule in Calculus to differentiate both sides of the equation concerning time. Differentiate both sides of the equation concerning time (or any other rate of change). Often, the chain rule is applied at this step.
  7. Substitute all known values into the resulting equation and solve for the required rate. Once done with the previous steps, it is now time to solve the wanted rate of change. Then, substitute all known values to get the final answer.

Note: A standard error is to substitute the given numerical information too early. It should be done only after the differentiation. Doing so will yield incorrect results since if used beforehand, those variables will become constants, and when differentiated, it would result in 0.

To fully understand these steps on how to do related rates, let us see the following word problems about associated rates.

Example 1: Related Rates Cone Problem

A water storage tank is an inverted circular cone with a base radius of 2 meters and a height of 4 meters. If water is being pumped into the tank at a rate of 2 m3 per minute, find the rate at which the water level rises when water is 3 meters deep.

Example 1: Related Rates Cone Problem

Example 1: Related Rates Cone Problem

Solution

We first sketch the cone and label it, as shown in the figure above. Let V, r, and h be the volume of the cone, the radius of the surface, and the water's height at time t, where t is measured in minutes.

We are given that dV/dt=2 m3/min, and we are asked to find dh/dt when the height is 3 meters. The quantities V and h are related by the formula of the cone's volume. See the equation shown below.

V = (1/3) πr2h

Remember that we want to find the change in height concerning time. Hence, it is very beneficial to express V as a function of h alone. To eliminate r, we use the similar triangles shown in the figure above.

r/h = 2/4

r = h/2

Substituting the expression for V becomes

V = 1/3π (h/2)2 (h)

V = (π/12) (h)3

Next, differentiate each side of the equation in terms of r.

dV/dt = (π/4) (h)2 dh/dt

dh/dt = (4/πh2) dV/dt

Substituting h=3 m and dV/dt=2m3/min, we have

dh/dt = (4/[π(3)2]) (2)
dh/dt = 8/9π

Final Answer

The water level is rising at a rate of 8/9π ≈ 0.28m/min.

Example 2: Related Rates Shadow Problem

A light is on top of a 15 feet tall pole. A 5 feet 10 inches tall person walks away from the light pole at a rate of 1.5 feet/second. At what pace is the shadow's tip moving out when the person is 30 feet from the bar pole?

Example 2: Related Rates Shadow Problem

Example 2: Related Rates Shadow Problem

Solution

Let us start by sketching the diagram based on the provided information from the problem.

Let x be the distance of the shadow's tip from the pole, p be the person's distance from the bar pole, and s be the shadow's length. Also, convert the person's height to feet for uniformity and more comfortable solving. The converted height of the person is 5ft 10 in = 5.83 feet.

The tip of the shadow is defined by the rays of light just getting past the person. Observe that they form a set of similar triangles.

Given the provided information and the unknown, relate these variables into one equation.

x = p + s

Eliminate s from the equation and express the equation in terms of p. Use the similar triangles shown from the figure above.

5.83/15 = s/x

s = (5.83/15) (x)

x = p + s

x = p + (5.83/15) (x)

p = (917/1500) (x)

x = (1500/917) (p)

Differentiate each side and solve for the required related rate.

dx/dt = (1500/917) (dp/dt)

dx/dt = (1500/917) (1.5)

dx/dt = 2.454 feet/second

Final Answer

The shadow's tip is then moving away from the pole at a rate of 2.454 ft/sec.

Example 3: Related Rates Ladder Problem

A ladder 8 meters long rests against a vertical wall of a building. The bottom of the ladder slides away from the wall at a rate of 1.5 m/s. How fast is the top of the ladder sliding down when the bottom of the ladder is 4 m from the building wall?

Example 3: Related Rates Ladder Problem

Example 3: Related Rates Ladder Problem

Solution

We first draw a diagram to visualize the ladder sitting against the vertical wall. Let x meters be the horizontal distance from the bottom of the ladder to the wall and y meters the vertical distance from the top of the ladder to the ground line. Note that x and y are functions of time, which is measured in seconds.

We are given that dx/dt = 1.5 m/s and we are asked to find dy/dt when x = 4 meters. In this problem, the relationship between x and y is given by the Pythagorean Theorem.

x2 + y2 = 64

Differentiate each side in terms of t using the chain rule.

2x (dx/dt) + 2y (dy/dt) = 0

Solve the previous equation for the desired rate, which is dy/dt; we obtain the following:

dy/dt = −x/y (dx/dt)

When x = 4, the Pythagorean Theorem gives y = 4√3, and so, substituting these values and dx/dt = 1.5, we have the following equations.

dy/dt = −(3/4√3)(1.5) = − 0.65 m/s

The fact that dy/dt is negative means that the distance from the top of the ladder to the ground decreases at a rate of 0.65 m/s.

Final Answer

The top of the ladder is sliding down the wall at a rate of 0.65 meters/second.

Example 4: Related Rates Circle Problem

Crude oil from an unused well is diffusing outward in the form of a circular film on groundwater's surface. If the radius of the circular film is increasing at the rate of 1.2 meters per minute, how fast is the area of the oil film spreading at the instant when the radius is 165 m?

Example 4: Related Rates Circle Problem

Example 4: Related Rates Circle Problem

Solution

Let r and A be the radius and area of the circle, respectively. Take note that the variable t is in minutes. The rate of change of the oil film is given by the derivative dA/dt, where

A = πr2

Differentiate both sides of the area equation using the chain rule.

dA/dt = d/dt (πr2)=2πr (dr/dt)

It is given dr/dt = 1.2 meters/minute. Substitute and solve for the growing rate of the oil spot.

(2πr) dr/dt = 2πr (1.2) = 2.4πr

Substitute the value of r = 165 m to the obtained equation.

dA/dt = 1244.07 m2/min

Final Answer

The oil film area growing at the instant when the radius is 165 m is 1244.07 m2/min.

Example 5: Related Rates Cylinder

A cylindrical tank with a radius of 10 m is being filled with treated water at a rate of 5 m3/min. How fast is the height of the water increasing?

Example 5: Related Rates Cylinder

Example 5: Related Rates Cylinder

Solution

Let r be the cylindrical tank's radius, h be the height, and V be the cylinder's volume. We are given a radius of 10 m, and the tank's rate is being filled with water, which is five m3/min. So, the volume of the cylinder is provided by the formula below. Use the volume formula of the cylinder to relate the two variables.

V = πr2h

Implicitly differentiate each side using the chain rule.

dV/dt = 2πr (dh/dt)

It is given dV/dt = 5 m^3/min. Substitute the given rate of change in volume and the tank's radius and solve the increase in height dh/dt of the water.

5 = 2π (10) (dh/dt)

dh/dt = 1/4π meter/minute

Final Answer

The height of water in the cylindrical tank is increasing at the rate of 1/4π meter/minute.

Example 6: Related Rates Sphere

Air is being pumped into a spherical balloon so that its volume increases at a rate of 120 cm3 per second. How fast is the radius of the balloon increasing when the diameter is 50 centimeters?

Example 6: Related Rates Sphere

Example 6: Related Rates Sphere

Solution

Let's start by identifying the given information and the unknown. The rate of increase in the volume of air is given as 120 cm3 per second. The unknown is the rate of growth in the sphere's radius when the diameter is 50 centimeters. Refer to the given figure below.

Let V be the volume of the spherical balloon and r be its radius. The rate of increase in volume and the rate of increase in radius may now be written as:

dV/dt = 120 cm3/s

dr/dt when r = 25cm

To connect dV/dt and dr/dt, we first relate V and r by the formula for the sphere's volume.

V = (4/3)πr3

To use the given information, we differentiate each side of this equation. To get the derivative of the right side of the equation, utilize the chain rule.

dV/dt = (dV/dr) (dr/dt) = 4πr2 (dr/dt)

Next, solve for the unknown quantity.

dr/dt = 1/4πr2 (dV/dt)

If we put r = 25 and dV/dt = 120 in this equation, we obtain the following results.

dr/dt = (1/[4π(25)2]) (120) = 6/(125π)

Final Answer

The spherical balloon radius is increasing at the rate of 6/(125π) ≈ 0.048 cm/s.

Example 7: Related Rates Travelling Cars

Car X is travelling west at 95 km/h, and car Y is travelling north at 105 km/h. Both cars X and Y are headed for the intersection of the two roads. At what rate are the cars approaching each other when car X is 50 m, and car Y is 70 m from the intersections?

Example 7: Related Rates Travelling Cars

Example 7: Related Rates Travelling Cars

Solution

Draw the figure and make C the intersection of the roads. At a given time of t, let x be the distance from car A to C, let y be the distance from car B to C, and let z be the distance between the cars. Take note that x, y, and z are measured in kilometers.

We are given that dx/dt= - 95 km/h and dy/dt = -105 km/h. As you can observe, the derivatives are negative. It is because both x and y are decreasing. We are asked to find dz/dt. The Pythagorean Theorem gives the equation that relates x, y, and z.

z2=x2+y2

Differentiate each side using the Chain Rule.

2z (dz/dt) = 2x (dx/dt) + 2y (dy/dt)

dz/dt = (1/z) [x (dx/dt) + y (dy/dt)]

When x = 0.05 km and y = 0.07 km, the Pythagorean Theorem gives z=0.09 km, so

dz/dt = 1/0.09 [0.05(−95) + 0.07(−105)]

dz/dt = −134.44 km/h

Final Answer

The cars are approaching each other at a rate of 134.44 km/h.

Example 8: Related Rates with Angles of Searchlight

A man walks along a straight path at a speed of 2 m/s. A searchlight is located on the floor 9 m from the straight path and is concentrated on the man. At what rate is the searchlight revolving when the man is 10 m from the point on the straightway closest to the searchlight?

Example 8: Related Rates with Angles of Searchlight

Example 8: Related Rates with Angles of Searchlight

Solution

Draw the figure and let x be the distance from the man to the point on the path closest to the searchlight. We allow θ be the angle between the ray of the searchlight and the perpendicular to the course.

We are given that dx/dt = 2 m/s and are asked to find dθ/dt when x = 10. The equation that relates to x and θ can be written from the figure above.

x/9 = tanθ

x = 9tanθ

Differentiating each side using implicit differentiation, we get the following solution.

dx/dt = 9sec2(θ) dθ/dt

dθ/dt = (1/9) cos2(θ) dxdt

dθ/dt = 1/9 cos2θ(2) = 2/9cos2(θ)

When x = 10, the beam's length is √181, so cos(θ)=9/√181.

dθ/dt = (2/9)( 9/√181)2 = (18/181) = 0.0994

Final Answer

The searchlight is rotating at a rate of 0.0994 rad/s.

Example 9: Related Rates Triangle

A triangle has two sides a = 2 cm and b = 3 cm. How fast is the third side c increasing when the angle α between the given sides is 60° and is expanding at the rate of 3° per second?

Example 9: Related Rates Triangle

Example 9: Related Rates Triangle

Solution

According to the law of cosines,

c2 = a2+b2 − 2ab(cosα)

Differentiate both sides of this equation.

(d/dt)(c2) = (d/dt)(a2 + b2 − 2abcosα)

2c (dc/dt) = −2ab(−sinα) dα/dx

dc/dt = [(absinα)/c] (dα/dt)

Calculate the length of the side c.

c = √(a2+b2−2abcosα)

c = √(22 + 32 − 2(2)(3)cos60°)

c = √7

Solve for the rate of change dc/dt.

dc/dt = (absinα)/c (dα/dt)

dc/dt = ((2)(3)sin60°)/ √7 (dα/dt)

dc/dt = ((2)(3)sin60°)/ √7 (3)

dc/dt = 5.89 cm/sec

Final Answer

The third side c is increasing at a rate of 5.89 cm/sec.

Example 10: Related Rates Rectangle

The length of a rectangle is increasing at a rate of 10 m/s and its width at 5 m/s. When the length measure is 25 meters and the width is 15 meters, how fast is the area of the rectangular section increasing?

Example 10: Related Rates Rectangle

Example 10: Related Rates Rectangle

Solution

Imagine the look of the rectangle to solve. Sketch and label the diagram as shown. We are given that dl/dt = 10 m/s and dw/dt = 5 m/s. The equation that relates the rate of change of the sides to the area is given below.

A = lw

Solve for the derivatives of the area equation of the rectangle using implicit differentiation.

d/dt (A) = d/dt (lw)

dA/dt = l (dw/dt) + w (dl/dt)

Use the given values of dl/dt and dw/dt to the obtained equation.

dA/dt = l (dw/dt) + w (dl/dt)

dA/dt = (25) (5) + (15) (10)

dA/dt = 275 m2/s

Final Answer

The area of the rectangle is increasing at a rate of 275 m2/s.

Example 11: Related Rates Square

The side of a square is increasing at a rate of 8 cm2/s. Find the enlargement rate of its area when the area is 24 cm2.

Example 11: Related Rates Square

Example 11: Related Rates Square

Solution

Sketch the situation of the square described in the problem. Since we are dealing with an area, the primary equation must be the square’s area.

A = s2

Implicitly differentiate the equation and take its derivative.

d/dt [A] = d/dt [s2]

dA/dt = 2s (ds/dt)

Solve for the measure of the square’s side, given the A = 24 cm2.

24 cm2 = s2

s = 2√6 cm

Solve for the required rate of change of the square. Substitute the value of ds/dt = 8 cm2/s and s = 2√6 cm to the obtained equation.

dA/dt = 2(2√6)(8)

dA/dt = 32√6 cm2/s

Final Answer

The area of the given square is increasing at a rate of 32√6 cm2/s.

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This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.

© 2020 Ray