# How to Solve Related Rates Problems in Calculus 1

• MHB
• harpazo
In summary, this problem involves related rates and similar triangles. The proportion 15/6= (x+ y)/x is used to find the distance of the man from the building and the length of his shadow. The final answer is that the shadow is changing at a rate of 10/3 ft/sec or 3.33 feet per second.
harpazo
I decided to review a few calculus 1 topics of interest. I like related rates but setting up the proper equation has been a big problem for me.

Question:

A light is on top of a building that is 15 ft. high. A man, 6 ft. tall, is walking away from the building at the rate of 2 ft/sec. At what rate is the shadow of the man changing when he is 4 ft. away from the building?

My Work:

This is a related rates problem given by the hint "At what rate is the shadow of the man changing..."

I think a similar triangle set up is needed here.

(Height of building)/(height of man) = y/(y - x), where y is the distance from the building to the man and (y - x) is the distance of the man's shadow.

NOTE: SHOULD THE DENOMINATOR BE (y - x) or (y + x)?
What's the difference?

The proportion then becomes

15/6 = y/(y - x)

Solving for y, I get y = (5x/3).

I now must differentiate both sides with
respect to time t, then substitute in what I know for the values of the variables. Since I want the speed of the shadow, I want to find out what dy/dt equals.

d/dt [y] = d/dt [(5x/3)

dy/dt = (dx/dt)(5/3)

I am given that dx/dt = 2 feet per second.

dy/dt = 2(5/3)

dy/dt = 10/3 ft/sec

Can I also write the answer as

dy/dt = 3.33 feet per second?

Is any of this correct?

Harpazo said:
I decided to review a few calculus 1 topics of interest. I like related rates but setting up the proper equation has been a big problem for me.

Question:

A light is on top of a building that is 15 ft. high. A man, 6 ft. tall, is walking away from the building at the rate of 2 ft/sec. At what rate is the shadow of the man changing when he is 4 ft. away from the building?

My Work:

This is a related rates problem given by the hint "At what rate is the shadow of the man changing..."

I think a similar triangle set up is needed here.

(Height of building)/(height of man) = y/(y - x), where y is the distance from the building to the man and (y - x) is the distance of the man's shadow.

NOTE: SHOULD THE DENOMINATOR BE (y - x) or (y + x)?
What's the difference?
You are correct that this involves two similar triangles. The shadow itself extends from the man in the direction away from the light. The larger of the two triangles has "height of the building" as one side and "distance to the end of the man's shadow" as another. The smaller triangle has the "height of man" as side corresponding to "height of the building" and "length of the triangle" as the side corresponding to "distance to the end of the man's shadow". The "distance from the building to the man", that you have called "y", is not part of either triangle. If you call "the distance from the building to the man" y, and the "length of the shadow" x, then the side of the large triangle corresponding to the height of the building is x+ y and the side of the triangle corresponding to the height of the man is x.

That gives the proportion 15/6= (x+ y)/x.

The proportion then becomes

15/6 = y/(y - x)

Solving for y, I get y = (5x/3).

I now must differentiate both sides with
respect to time t, then substitute in what I know for the values of the variables. Since I want the speed of the shadow, I want to find out what dy/dt equals.

d/dt [y] = d/dt [(5x/3)

dy/dt = (dx/dt)(5/3)

I am given that dx/dt = 2 feet per second.

dy/dt = 2(5/3)

dy/dt = 10/3 ft/sec

Can I also write the answer as

dy/dt = 3.33 feet per second?

Is any of this correct?

HallsofIvy said:
You are correct that this involves two similar triangles. The shadow itself extends from the man in the direction away from the light. The larger of the two triangles has "height of the building" as one side and "distance to the end of the man's shadow" as another. The smaller triangle has the "height of man" as side corresponding to "height of the building" and "length of the triangle" as the side corresponding to "distance to the end of the man's shadow". The "distance from the building to the man", that you have called "y", is not part of either triangle. If you call "the distance from the building to the man" y, and the "length of the shadow" x, then the side of the large triangle corresponding to the height of the building is x+ y and the side of the triangle corresponding to the height of the man is x.

That gives the proportion 15/6= (x+ y)/x.

This is a related rates problem given by the hint "At what rate is the shadow of the man changing..."

We have a similar triangle set up to deal with here.

The set up should be

the proportion 15/6= (x+ y)/x.

Solve for y.

6(x + y) = 15x

6x + 6y = 15x

6y = 15x - 6x

6y = 9x

y = 9x/6

y = 3x/2

Differentiate both sides.

d/dt [y] = d/dt [(3x/2)]

dy/dt = (dx/dt)(3/2).

We are given that dx/dt = 2 feet per second.

We need to solve for dy/dt.

dy/dt = 2(3/2)

dy/dt = 3 feet per second is the correct answer.

Suppose we wanted to work the problem in general terms. Please refer to the following diagram:

View attachment 6439

$P$ is the height of the pole, $M$ is the height of the man, $D$ is the man's distance from the pole, and $x$ is the length of the shadow.

By similarity, we find:

$$\displaystyle \frac{D+x}{P}=\frac{x}{M}$$

$$\displaystyle DM+Mx=Px$$

$$\displaystyle x=\frac{DM}{P-M}$$

Observing that $x$ and $D$ are changing with time, we find by implicitly differentiating with respect to time $t$, we have:

$$\displaystyle \frac{dx}{dt}=\frac{M}{P-M}\frac{dD}{dt}$$

And so, using the given data, we find:

$$\displaystyle P=15\text{ ft},\,M=6\text{ ft},\,\frac{dD}{dt}=2\,\frac{\text{ft}}{\text{s}}$$

$$\displaystyle \frac{dx}{dt}=\frac{6}{15-6}\cdot2\,\frac{\text{ft}}{\text{s}}=\frac{4}{3}\,\frac{\text{ft}}{\text{s}}$$

If we wanted to find how fast the tip of the shadow is moving, then we observe that movement of the tip of the man's shadow is the result not only of the shadow growing but also of the man's movement, we find that we must add these rates of change, so that the rate of change of the point $T$ of the tip of the shadow is given by:

$$\displaystyle \frac{dT}{dt}=\frac{dD}{dt}+\frac{dx}{dt}=\frac{dD}{dt}\left(1+\frac{M}{P-M} \right)=\frac{P}{P-M}\frac{dD}{dt}$$

Now, using the given data (did you notice we do not need $D$?):

$$\displaystyle P=15\text{ ft},\,M=6\text{ ft},\,\frac{dD}{dt}=2\,\frac{\text{ft}}{\text{s}}$$

We find:

$$\displaystyle \frac{dT}{dt}=\frac{15}{15-6}\cdot2\,\frac{\text{ft}}{\text{s}}=\frac{10}{3}\, \frac{\text{ft}}{\text{s}}$$

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MarkFL said:
Suppose we wanted to work the problem in general terms. Please refer to the following diagram:
$P$ is the height of the pole, $M$ is the height of the man, $D$ is the man's distance from the pole, and $x$ is the length of the shadow.

By similarity, we find:

$$\displaystyle \frac{D+x}{P}=\frac{x}{M}$$

$$\displaystyle DM+Mx=Px$$

$$\displaystyle x=\frac{DM}{P-M}$$

Observing that $x$ and $D$ are changing with time, we find by implicitly differentiating with respect to time $t$, we have:

$$\displaystyle \frac{dx}{dt}=\frac{M}{P-M}\frac{dD}{dt}$$

And so, using the given data, we find:

$$\displaystyle P=15\text{ ft},\,M=6\text{ ft},\,\frac{dD}{dt}=2\,\frac{\text{ft}}{\text{s}}$$

$$\displaystyle \frac{dx}{dt}=\frac{6}{15-6}\cdot2\,\frac{\text{ft}}{\text{s}}=\frac{4}{3}\,\frac{\text{ft}}{\text{s}}$$

If we wanted to find how fast the tip of the shadow is moving, then we observe that movement of the tip of the man's shadow is the result not only of the shadow growing but also of the man's movement, we find that we must add these rates of change, so that the rate of change of the point $T$ of the tip of the shadow is given by:

$$\displaystyle \frac{dT}{dt}=\frac{dD}{dt}+\frac{dx}{dt}=\frac{dD}{dt}\left(1+\frac{M}{P-M} \right)=\frac{P}{P-M}\frac{dD}{dt}$$

Now, using the given data (did you notice we do not need $D$?):

$$\displaystyle P=15\text{ ft},\,M=6\text{ ft},\,\frac{dD}{dt}=2\,\frac{\text{ft}}{\text{s}}$$

We find:

$$\displaystyle \frac{dT}{dt}=\frac{15}{15-6}\cdot2\,\frac{\text{ft}}{\text{s}}=\frac{10}{3}\, \frac{\text{ft}}{\text{s}}$$

Are you saying that my original aswer is right? I am really lost now.

Harpazo said:
Are you saying that my original aswer is right? I am really lost now.

I got 4/3 ft/s where you got 3 ft/s.

You have the line:

$$\displaystyle \d{y}{t}=\frac{3}{2}\d{x}{t}$$

Now, this implies:

$$\displaystyle \d{x}{t}=\frac{2}{3}\d{y}{t}$$

And from this, you get the answer I gave. :D

MarkFL said:
I got 4/3 ft/s where you got 3 ft/s.

You have the line:

$$\displaystyle \d{y}{t}=\frac{3}{2}\d{x}{t}$$

Now, this implies:

$$\displaystyle \d{x}{t}=\frac{2}{3}\d{y}{t}$$

And from this, you get the answer I gave. :D

Originally, I got (10/3) ft/sec as the answer. Let me get this straight. The answer is (4/3) ft/sec, right?

Harpazo said:
Originally, I got (10/3) ft/sec as the answer. Let me get this straight. The answer is (4/3) ft/sec, right?

Yes, the length of his shadow is increasing at a rate of 4/3 ft/s, and the tip of his shadow is moving at a rate of 10/3 ft/s. :D

MarkFL said:
Yes, the length of his shadow is increasing at a rate of 4/3 ft/s, and the tip of his shadow is moving at a rate of 10/3 ft/s. :D

Ok. My answer is correct for a different question. If we are seeking the rate at which the tip of the man's shadow is moving, then (10/3) ft/sec is correct.

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## 1. What are related rates problems in Calculus 1?

Related rates problems in Calculus 1 involve finding the rate of change of one variable with respect to another variable in a given scenario. These types of problems often involve multiple variables that are related to each other through a mathematical equation or model.

## 2. How do I approach solving related rates problems in Calculus 1?

The first step in solving related rates problems is to identify the variables involved and their relationship to each other. Then, use the given information to set up an equation that represents the relationship between the variables. Finally, take the derivative of both sides of the equation and plug in the given values to solve for the unknown rate of change.

## 3. What are some common types of related rates problems in Calculus 1?

Some common types of related rates problems include geometric problems involving changing dimensions, motion problems involving changing velocity or acceleration, and volume or area problems involving changing rates of change.

## 4. How do I know when to use the chain rule in related rates problems?

The chain rule is used in related rates problems when the variables involved are changing with respect to different independent variables. This often happens when the variables are related through multiple equations or when the problem involves a moving object.

## 5. What are some tips for solving related rates problems in Calculus 1?

Some helpful tips for solving related rates problems include carefully reading and understanding the given information, drawing a diagram or visual representation of the problem, and setting up and solving the equations step by step. It is also important to check the units of the given variables and the final answer to ensure they are consistent.

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