Asked by Pat
A 5ft person is running away from a 50ft light post at 13ft/s. Find the rate of change of the shadow from the person when they are 60 feet from the light post.
So I made a right triangle diagram, where the vertical side is 50ft, the bottom side is x + y feet (x = shadow length, y = 60ft), and inside the triangle is a 5ft vertical line where x and y meet.
I know that dy/dt is 13ft/s. The ROC of the distance between the person and light post is increasing, so it is positive.
I know that I want to find dx/dt, the ROC of the shadow length when y = 60ft.
So I did similar triangles: 5/x = 50/(x+y) --> 5x + 5y = 50x --> y = 9x
And then differentiated and substituted dy/dx with the ROC: 13 = 9*(dx/dt)
I got dx/dt = 13/9 ft/s.
However, I did not do anything with y = 60ft. This dx/dt didn't really tell me what the ROC is when y = 60ft. I need some help.
So I made a right triangle diagram, where the vertical side is 50ft, the bottom side is x + y feet (x = shadow length, y = 60ft), and inside the triangle is a 5ft vertical line where x and y meet.
I know that dy/dt is 13ft/s. The ROC of the distance between the person and light post is increasing, so it is positive.
I know that I want to find dx/dt, the ROC of the shadow length when y = 60ft.
So I did similar triangles: 5/x = 50/(x+y) --> 5x + 5y = 50x --> y = 9x
And then differentiated and substituted dy/dx with the ROC: 13 = 9*(dx/dt)
I got dx/dt = 13/9 ft/s.
However, I did not do anything with y = 60ft. This dx/dt didn't really tell me what the ROC is when y = 60ft. I need some help.
Answers
Answered by
oobleck
The distance does not matter.
The ROC of the shadow is constant, just like dy/dt
The ROC of the shadow is constant, just like dy/dt
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