How to Solve Related Rates Problems in Calculus 1

[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?

 

[HallsofIvy]

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?

 

[harpazo]

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.

 

[MarkFL]

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:

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

 

[harpazo]

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.

 

[MarkFL]

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

 

[harpazo]

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?

 

[MarkFL]

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

 

[harpazo]

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.

- - - Updated - - -