A man 6ft tall walks at a rate of 5ft/s away from a lampost that is 19ft high. At what rate is the length of?

Question

his shadow changing when he is 35ft away from the lampost?
*please explain steps; any help is greatly appreciated**
possible answers:
a. 6/5
b. 30/13
c. 3/5
d. 175/6

Answer 1

a sq + b sq= c sq

Answer 2

by comparable triangles if H be the submit top H/18 = 6/10 = 3/5 H = fifty 4/5 = 10.8 ft enable D+S = distance from submit and D+S = shadow length 10.8/[72ff98d875cd479b5f8629473b6028072ff98d875cd479b5f8629473b6028072ff98d875cd479b5f8629473b60280] = 6/D+S 10.872ff98d875cd479b5f8629473b60280=672ff98d875cd479b5f8629473b6028072ff98d875cd479b5f8629473b60280672ff98d875cd479b5f8629473b60280 D+S = a million.2572ff98d875cd479b5f8629473b60280 d72ff98d875cd479b5f8629473b60280/dt = a million.25d72ff98d875cd479b5f8629473b60280/dt = a million.25*50 = sixty two.5 ft/min

Answer 3

We first need to find a formula that relates the distance of the man from the lamppost with the length of the shadow.
I’m sorry I can’t draw this for you.

I’m going to form a triangle from the bottom of the lamp post to the top of the lamppost to the end of the man’s shadow. The man is within the triangle with his head touching the hypotenuse. I’m going to use x to represent the distance from the man to the lamppost, and y to represent the length of the shadow.

From geometry, we know that the ratio of the lamppost height to the length from the lamppost to the end of the shadow is the same as the height of the man to the length of the shadow.

19 / (x + y) = 6 / y
19y = 6x + 6y
13y = 6x
Take d/dt on both sides
13(dy/dt) = 6(dx/dt)
You know that dx/dt = 5 ft/s
13(dy / dt) = 6(5 ft/s)
dy / dt = 30/13 ft/s

So no matter what distance he is away from the lamppost, as long as he is constantly walking at 5 ft/s his shadow will always grow at 30/13 ft/s.

Answer 4

Assuming the source of light is the top of the post?

Note the similar triangles. Write equation for length of smaller triangle's bottom edge as a function of time.

Now apply the definition of derivative and plug in the value for distance from the pole at t=7.