Solve this problem without using a similar triangles approach: 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?

Question

hero · Answer

Draw it. There's drawing function below

hero · Answer

If you can't use similar triangles, what approach are you supposed to use?

anonymous · Answer

is there a way to find it without using it?

hero · Answer

Not really sure. Why?

anonymous · Answer

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well that's what im trying to find. i understand it using similar triangles, but im trying to find another approach.

hero · Answer

Why would you want to use another approach? What's wrong with the one you already have?

anonymous · Answer

i tried breaking it up using the area of a triangle and adding it to the area of a trapazoid, to the find the area of total. however, when i took the derivative, i had too many missing variables, including dy/dt.
that's the question im being asked.

anonymous · Answer

$$A _{total} = h ((b _{1} + b _{2}) / 2) + (1/2 * bh)$$
i tried using this. when i plugged what i had into the derivative of this, it looked like this:
$$d/d _{t} = 21/2 *5 + 3*d _{y}/d _{t}$$
dx/dt was 5ft/s, the given speed of the man.

does this work?

hero · Answer

Ask Satellite

hero · Answer

Okay, I can get you an answer but it will take a little time

hero · Answer

But I can't promise that the solution will not include similar triangles

anonymous · Answer

oh ok. thanks. 
instead of using that equation i gave above, i just used the velocity formula knowing that it's the derivative of an equation.
$$V = x / t$$
where it becomes in terms of d/dt:
$$dV/dt = dx/dt \div d/dt$$
substitute what is given, and i got:
$$d/dt = 40/5$$
$$d/dt = 8ft/s$$

the answer in my book is 25/3 ft/s, which equals roughly 8.3 ft/s