- #1
student34
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Homework Statement
(I don't know how to draw a picture here so I will just explain the dimensions of the similar triangles)
A lightbulb on a floor shines at a 2m tall guy who is walking away from it towards a wall. As he gets closer to the wall, his shadow gets smaller and smaller on the wall. The lightbulb is 10m from the bottom of the wall. This makes a triangle where the base is 10m and the height of it is the height of the shadow (y). At what rate is the height of the shadow decreasing if the guy walks at a rate of 1.2m/s and is 7m (we will call x) from the lightbulb. This information creates a similar small triangle with a base of x = 7m (his distance from the lightbulb) and a height (his height) of 2m.
The Attempt at a Solution
So, in my notes my instructor found the related rate by using a similar triangle method of y/10 = 2/x When differentiated it looks like y' = -20/x^2(x') = -20/49(1.2) = -(24/49)m/s. I was fine with this, and it makes sense to me until I tried a different similar triangle method. I tried putting y/(x+3) = 2/7 which gives me the correct x and y dimensions of the triangles. But when I differentiate the equation, I don't get the right answer. Instead, I get:
y = (2x)/7 + 3/7
y' = (2(x'))/7 + 0
y' = (2(1.2))/7
y' = 12/35 which has the wrong sign and is a constant change - not an accelerated one.
I never want to stop bashing my head into the wall until I understand why this doesn't work.