Related rates problem

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Darya

Junior Member
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Jan 17, 2020
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154
A street light is mounted at the top of a 15-ft-tall pole. A man
6 ft tall walks away from the pole with a speed of 5 ft!s along
a straight path. How fast is the tip of his shadow moving when
he is 40 ft from the pole?

Here's my work. I keep getting 10/3 but the correct answer is 25/3. I also re-did it using quotient rule but it's the same result. Where's the mistake?

Thanks!!!
Thaphoto_2020-04-03_14-26-28.jpg
 
I'd say you need to let x be as you have it, but y be the distance from the tip of the shadow back to the base of the flagpole.

Your equation then becomes 156=yy−x\displaystyle \frac{15}{6} =\frac{y}{y-x} which leads to y=53x\displaystyle y=\frac{5}{3}x
 
To add to the response above:

According to your set-up, the 'y' measurement will have a moving origin.

In your set-up you are measuring - how fast the tip of your shadow moving away from you - not from a fixed origin like the pole.
 
Think about what is happening to the tip of the shadow as the man walks away from the street light.

From your diagram the rate the tip moves is the rate he walks away from the light PLUS the rate
from where he is standing to the tip of the shadow. You are given the rate he walks away from
the light: dx/dt = 5 ft/s. The rate between the man and the tip of the shadow is dy/dt=10/3 ft/s.

The rate the tip of the shadow is moving is the sum of the two rates.

dx/dt + dy/dt = 5 + (10/3 )= (15/3)+(10/3)=(25/3) ft/sec
 
Li Batton said:
Think about what is happening to the tip of the shadow as the man walks away from the street light.

From your diagram the rate the tip moves is the rate he walks away from the light PLUS the rate
from where he is standing to the tip of the shadow. You are given the rate he walks away from
the light: dx/dt = 5 ft/s. The rate between the man and the tip of the shadow is dy/dt=10/3 ft/s.

The rate the tip of the shadow is moving is the sum of the two rates.

dx/dt + dy/dt = 5 + (10/3 )= (15/3)+(10/3)=(25/3) ft/sec
Click to expand...
OHHHHH got it!!!!!
I'm really thankful!!
 
Your last line is wrong! Terribly wrong. But that doesn't matter here.

You can define y to be whatever you want. The problem is that you need to decide what you want to solve for. You want to find d(x+y)/dt since the length of the shadow is x+y !!

Since (x+y) = (15/6)y and y = (2/3)x we have (x+y) = (5/3)x. Then d(x+y)/dt = (5/3)dx/dt = (5/3)(5) = 25/3.
 
Jomo said:
Your last line is wrong! Terribly wrong. But that doesn't matter here.

You can define y to be whatever you want. The problem is that you need to decide what you want to solve for. You want to find d(x+y)/dt since the length of the shadow is x+y !!

Since (x+y) = (15/6)y and y = (2/3)x we have (x+y) = (5/3)x. Then d(x+y)/dt = (5/3)dx/dt = (5/3)(5) = 25/3.
Click to expand...
Yes, you are right. I then realized I should've looked for d(x+y)/dt
But why is the last line wrong? Can't we subtract dy/dt from 5/6(dy/dt)?
 
Darya said:
Yes, you are right. I then realized I should've looked for d(x+y)/dt
But why is the last line wrong? Can't we subtract dy/dt from 5/6(dy/dt)?
Click to expand...
Actually I thought it was wrong, then realized it was correct and when I made my post I forget that it was right!! I am sorry about that. You work was all fine except for the fact that you solved for the wrong derivative.
 
Jomo said:
Actually I thought it was wrong, then realized it was correct and when I made my post I forget that it was right!! I am sorry about that. You work was all fine except for the fact that you solved for the wrong derivative.
Click to expand...
Now go to the corner and contemplate on your blunder (I mean mistake) for 25/3 minutes!!!
 
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