That's because you found the rate at which the shadow length is changing.
The tip of the shadow is x+y feet from the light. Therefore, the rate it is
changing is dx/dt+dy/dt. You have 0.8 + 0.186 = 0.986
I posted a question almost the same as this yesterday.. and I understand how to do them. I followed the way the people who answered my post solved the problem, but it's not working for this question.
A 1.7 m tall man walks away from a 9-m high streetlight at 0.8 m/s. How fast is the end of his shadow moving when he is 20 m from the lamppost.
My textbook says the answer is 0.986 m/s, but I keep on getting 0.186 m/s.
PLEASE HELP, i have a test tommorrow, and I know a question like this is going to be on it
Hello, jmailloux!
You did it again . . . you made the wrong part.
A 1.7-m tall man walks away from a 9-m high streetlight at 0.8 m/s.
How fast is the end of his shadow moving when he is 20 m from the lamppost?Code:* | * | * 9 | | * | 1.7| * | x | s-x * ----*-----------+-----------------------*---- : - - - - - - - - s - - - - - - - - :
This time, we make be the entire distance,
. . from the lampost to the end of his shadow.
From the similar right triangles, we have: .
Then: .
Differentiate with respect to time: .
Since