Math question!

[batskeets]

This is James' question, actually, but I don't remember how to do it, either, so here it is:

A street light is mounted at the top of a 15-foot-tall pole. A man 6 ft. tall walks away from the pole with a speed of 5 ft/second along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

James knows how to do it with partial derivatives, but he needs a reminder of how to do it with dx/dt and dy/dt. And as your teachers always say, show your work! (if you can)

I have no idea if any of you guys know this, but if'n you do, shout out! :) Thaaaank you!

EDIT: Never mind--he woke up today, and the realization of how to do the problem struck him about 10 minutes after being awake. (and this was, of course, after I suggested last night that he get some sleep and try it in the morning when his mind was fresh--I may not recall my calculus so well, but I still give some sensible advice!)

Comments

[threeleet.livejournal.com]

OK, no idea if this is anywhere near correct, but here goes:

A right trapezoid is created between the man & the street light, where the light post is 15', the man is 6', and the distance between them is x (this would be SO much easier w/ a diagram, but oh well). Drawing a straight line from the top of the man's head to the light post, parallel to the ground, gives a right triangle, so you can call the angle facing the ground θ. We know the height of the triangle is 9 (15-6), and the side parallel to the ground is x, so by trig, tan θ = x/9 (opposite/adjacent). So, if we extend the triangle down to the ground, the new corners are the top & bottom of the lamp, and the tip of the man's shadow. θ is still the same, and the height of the lamp is 15, so the distance from the lamp to the tip of the shadow (y) is found by the same equation: tan θ = y/15. Since both equations have tan θ, we can say x/9 = y/15, or y=15x/9.

Now, since x can also be stated in terms of time, we say that x is 5t. So y=75t/9. Thus, Δy =75Δt/9, or dy/dt = 15/9 dx/dt. IOW, dy/dt = 75/9, dx/dt = 5. Right?

[threeleet.livejournal.com]

How fast is the tip of his shadow moving when he is 40 ft from the pole?

Shit, I forgot to answer that part! Hmm.. the top paragraph is all fine, but the deltas & dy/dt stuff looks wrong... let's see.

At x=40, t=8, and y=66 2/3. x=5t, dx/dt=5dt. So, dy/dt = 75/9 dt.

Oh, OK: at t=8, y=66.6666, so the speed is Δy/Δt (= dy/dt?) = 66.666/8 = 8.33333 = 8 1/3 ft/s. Does that sound right?

[threeleet.livejournal.com]

Y'know, this is really bugging me: 66.6/8 = 8.333 = 75/9. IOW, the shadow is moving at a constant speed. Its speed at 40' from the pole is the same as its speed from 1'. Almost makes it sound too easy for the amount of work required... hmm.

And even if it was "how fast is his shadow GROWING at 40'", it'd still be a constant: dy - dx = 75dt/9 - 5dt = 8.333dt - 5dt = 3.333ft/s... Another constant. Why the hell are dy/dt & dx/dt even in this? Are they just trying to make it sound harder than it should be?

And how the bloody fuck did he manage to bring partial differentials into this?!

Alright, I've blown enough time on this... Damn you! ^_^

hey!

[threeleet.livejournal.com]

I just realized last night that you never told me if I was right or not! Damn you! And damn you for being eligible for the new photo hosting beta test. :P"

...And your little dog, too!

[kelaar.livejournal.com]

I missed that question on the exam. Seriously. Calculus didn't start to make sense until the end of spring term, and I've done none since then.

The answer is...

[randvek.livejournal.com]

It depends. The problem makes no mention of the street light being on, or if it is even night. It also only states that he is walking away from the pole, not starting at it. He could be down the street, walking away from the pole after eating at the local Denny's. I don't think the street light would be affecting his shadow much. However! Denny's is a 24 hour restaurant, so it might still be day or night, so the fact that he was eating at Denny's, or even eating breakfast at Denny's, does not help us in the least. I hate it when math problems give you information you don't need, making you wonder if you did something wrong when you didn't factor it in. Freaking Moons over My-Hammy is ruining this problem for me.

A puzzling puzzle indeed.

[muerte.livejournal.com]

42

[http://users.livejournal.com/axiom-/]

*the easy solution*

A) The angle the light makes with the mans head with respect to level ground is equal to the angle of the triangle the shadows tip makes with respect to the light. (Drawings make this obvious)

Light(15')
I_*
I____*
I_______*
I_________Man(6')
I_________I_*
I_________I____*
I_________I_______*
I_________Xm=f(t)____Xs

The tangent of the angle (the mans head with respect to level ground) is equal to the opposite over adjacent (15-6)/Xm, or height of pole minus height of man over the distance the man has traveled in t seconds.

Xm = 5f/s*t
Tan(angle)=(15-6)/5t

The tangent of the angle (Shadow tip to Light) is equal to the height of the pole over the position of the shadow (Xs).

Tan(angle)=15/Xs

These being equal angles

(15-6)/5t=15/Xs

Xs=(15*5t)/(15-6)
= 75/9*t

just differentiate with respect to t d/dt(Xs) = 75/9

Constant speed!