2nd Gen Help! Calculus Homework

Hey guys, My son is in calculus and he has a rates of change word problem. It has been a long time since I have done these haha!

I now realize how little I use certain maths when running a company.
Please help me out! I tried once but was wrong!

A street light is at the top of a 20 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 6 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 35 ft from the base of the pole?

 

He said a good tip was drawing a picture. I feel like the dumbest engineer ever

 

If you had asked me this time last year I would have known lol. I have since forgotten.

I know you need to make a function of some kind then take the derivative to get rate of change, then plug in 35'. I know that didn't help much, sorry.

 

6x = 20x-20b
-4x = -20b
x = 5b/2
we are given that her rate is 6 ft/s so db/dt=6. take the derivative of our function relating b to x with respect to time...
= Interesting note here is that her position relative to the pole is not even in this equation, that means that the rate her shadow moves is constant through her entire walk...
anyway...solve for dx/dt since we know db/dt= 6
dx/dt= (5/2) * 6 = 15 ft/s <<<< rate of her shadow when she is 35 feet from the light

 

What he said......

 

we ought to start a section for education/school stuff...lol

 

He plugged in 15 ft/sec on the Webwork and it was incorrect?! That looked legit

 

Don't feel bad, I haven't done problems like this in a looong time... My guess is 2.571 ft/s. I've got to say, I feel like my 4 years of engineering school has failed me big time...

 

Haha, nice man, guess I was way the hell off! good job!

 

That was apparently wrong, or the stupid online thing is wrong.

I seriously hate online homework lol.

 

i should know this! i am in calculus right now. what isu2014 looked about right. he has to find an equation that relates the info you know and the info you dont know.

did it give any other information besides what you provided us?

 

The 35 ft away from the pole does not matter. But 15 ft/sec was in correct.

 

http://www.physicsforums.com/forumdisplay.php?f=109

 

Did the 2.571 ft/s work? I'm not sure if it hurts you to try to check it, if it does I wouldn't try it but that's what I came up with in a round about sort of way.

 

Depends on whether or not it's daylight savings or not.

 

Good point, it never says the shadow is from the light pole, haha.

 

i took online math try 15.0 ft/sec

 

20/6=y/(y-x)
20y-20x=6y
y=20/9x

dx/dt=6

dy/dt=20/9*dx/dt
dy/dt=20/9*6=40/3 ft/sec


try that. i found a similar problem in my calc book and that i just adhusted the problem to fit your numbers

if you want me to try and explain it just let me know haha

 

So OP, what was the correct answer?