Related Rates - Sort of confused

[kevinr]

[SOLVED] Related Rates - Sort of confused

Homework Statement

There are two buildings. One 20' high and other 40' high with a 60' distance between them. A person walking between them creates a theta with the buildings (see pic)

PICTURE: http://allyoucanupload.webshots.com/v/2003959771064888674

Now i am asked to find maximum theta possible if the person is walking 4ft/s to the left.

Homework Equations

?

The Attempt at a Solution

I am kind of lost in coming up with the relation between theta and the two buildings. I know once i get the relation in terms of theta, i can take the derivative and find when the derivative = 0 and see if that's a max.

But any help with finding this equation would be greatly appreciated.

Thanks!

 

[HallsofIvy]

Are you sure that you have stated the problem correctly? The "maximum theta" will occur at a specific position no matter how fast the person is moving.

Let x be the distance the person is from the building on the right, in feet. Then the distance from the building on the left is 60- x.
Let $\phi$ be the angle the line from the top of the building on the left to the person makes with the ground. Then $tan(\phi)= 20/(60-x)$ so $\phi= arctan(20/(60-x))$.
Let $\psi$ be the angle the line from the top of the building on the right to the person makes with the ground. Then $tan(\psi)= 40/x$ so [itex]\psi= arctan(40/x).

Of course, $\theta$ is just $\pi- \phi- \psi= \pi- arctan(20/(60-x))- arctan(40/x)$. To find the x that gives the maximum value for $\theta$, differentiate that and set equal to 0.

Again, I don't see that how fast the person is walking has anything to do with that.

 

[kevinr]

o ok thx!