Related rates problem: A street light is mounted at the top of a 15'foot pole....

Linty Fresh

Junior Member
Joined
Sep 6, 2005
Messages
58
Hi all!

I'm having a bit of trouble with this related rates problem.

A street light is mounted at the top of a 15'foot pole. A man six feet 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?


OK, so I constructed the standard triangle with x being the distance of the tip of the man's shadow (x=40+s, "s" being the distance from the man to his shadow). "y" is the height of the streetlight, and "z" is the hypotenuse of the triangle (distance from the street light to the tip of the man's shadow).


First, I found the length of x. Using similar triangles:

6'/15' = s/(40+s) --> s=26.7 and x=66.7'. As y = 15', x^2 + y^2 = z^2, and z=68.4'


Deriving x^2 + y^2 = z^2, we get 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)


As dy/dt=0 (The street light does not move), we're left with 2x(dx/dt) = 2z(dz/dt)


Now I look at this, and it seems to me that dz/dt represents the speed of the man's shadow. dx/dt=5 ft/sec. Thus, solve for dz/dt, and I get 4.9 ft/sec. The book gives me 25/3 ft/sec.


What am I doing wrong? Many thanks!
 
Hi all!

I'm having a bit of trouble with this related rates problem.

A street light is mounted at the top of a 15'foot pole. A man six feet 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?


OK, so I constructed the standard triangle with x being the distance of the tip of the man's shadow (x=40+s, "s" being the distance from the man to his shadow). "y" is the height of the streetlight, and "z" is the hypotenuse of the triangle (distance from the street light to the tip of the man's shadow).


First, I found the length of x. Using similar triangles:

6'/15' = s/(40+s) --> s=26.7 and x=66.7'. As y = 15', x^2 + y^2 = z^2, and z=68.4'


Deriving x^2 + y^2 = z^2, we get 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)


As dy/dt=0 (The street light does not move), we're left with 2x(dx/dt) = 2z(dz/dt)


Now I look at this, and it seems to me that dz/dt represents the speed of the man's shadow.
No, it isn't. You said earlier "and "z" is the hypotenuse of the triangle (distance from the street light to the tip of the man's shadow)". Since you have "x=40+s, "s" being the distance from the man to his shadow", the length of the man's shadow is s, not z.



dx/dt=5 ft/sec. Thus, solve for dz/dt, and I get 4.9 ft/sec. The book gives me 25/3 ft/sec.


What am I doing wrong? Many thanks!
 
Yes, it is the rate, in feet per second, at which the length of the shadow is increasing.
 
Sorry, still a bit confused.


So to find ds/dt, should I go back to the Pythagorean theorem and frame it in terms of "s"? In other words, instead of "x," let "d" equal the distance from the light post to the man and write the equation as:


(d+s)^2+y^2=z^2


Derive both sides of this equation, and solve for ds/dt? I just feel as if I'm missing something obvious.



Thanks again.
 
Sorry, still a bit confused.


So to find ds/dt, should I go back to the Pythagorean theorem and frame it in terms of "s"? In other words, instead of "x," let "d" equal the distance from the light post to the man and write the equation as:


(d+s)^2+y^2=z^2


Derive both sides of this equation, and solve for ds/dt? I just feel as if I'm missing something obvious.



Thanks again.
Since x=40+s we get dx/dt = ds/dt. Now using your definition of the length finish the problem. If you need help, then tell us where you are stuck.
 
I think you have made this much more complex than it need be.

Using your definitions \(\displaystyle \dfrac{dx}{dt} = 5.\)

\(\displaystyle u = x + s.\)

You are asked to find \(\displaystyle \dfrac{du}{dt}.\)

\(\displaystyle \dfrac{6}{15} = \dfrac{s}{x + s} \implies 6x + 6s = 15s \implies s = \dfrac{2}{3} * x \implies\)

\(\displaystyle u = \dfrac{5}{3} * x \implies \dfrac{du}{dt} = \dfrac{5}{3} * \dfrac{dx}{dt} \implies \dfrac{du}{dt} = \dfrac{5}{3} * 5 = \dfrac{25}{3}.\)

If you are asked to find something, it is a good idea to give it its own variable.
 
Thanks, Jeff. You made it look so simple. I was too hung up on the Pythagorean theorem, and I didn't realize what the velocity of the shadow actually entailed. I do appreciate it!


On edit: Actually, looking back on it, my main problem was not realizing that to find the velocity of the shadow, I had to find the derivative of the entire base of the triangle (distance from light + length of shadow). I'm still just a bit hung up on that.
 
Last edited:
Thanks, Jeff. You made it look so simple. I was too hung up on the Pythagorean theorem, and I didn't realize what the velocity of the shadow actually entailed. I do appreciate it!


On edit: Actually, looking back on it, my main problem was not realizing that to find the velocity of the shadow, I had to find the derivative of the entire base of the triangle (distance from light + length of shadow). I'm still just a bit hung up on that.
Actually, I think you made two errors. I kept trying to make sense of your original post and getting all mixed up. It was only when I went back to the original problem and ignored your logic that I made progress.

Yes, you did not have clearly in mind what variable for which you were to find the rate of change with respect to time. I suspect you were groping for it, which is why I suggested naming a variable whose rate of change was being sought. But I think you at least knew that both x and s were relevant. You were close.

I think the other mistake, much more confusing, was not coming up with the general relationship between x and s. I think that mistake sent Jomo, who is a much better mathematician than I am, off into never-never land in this case.

\(\displaystyle \dfrac{6}{15} = \dfrac{s}{s + 40}\) is not generally true. It is true only if x = 40.

Once I realized that what is generally true is \(\displaystyle \dfrac{6}{15} = \dfrac{s}{x + s}\),

all became clear.
 
Top