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A 30-foot ladder rests vertically against a wall. A bug starts at the bottom of the ladder and climbs up at a rate of 3.5 feet per minute. At the same time, the foot of the ladder is being pulled along the ground at a rate of 1.5 feet per minute until the top of the ladder reaches the ground. Let $x$ be the distance of the bug from the wall at time $t$.

I got $x(t)=-.175t^2 +1.5t$ to represent the horizontal distance of the bug to the wall as a function of time, $t$.

The question also asks:

When the bug is farthest from the wall how high is it?

When the bug reaches the top of the ladder how high is it from the ground?

What speed does the bug need to climb in order to reach the top at the same time the top of the ladder reaches the ground?

What I am confused about is what equation/steps do I need to find in order to compare the equation I found with, which will allow me to find the answers. I looked up related rates problems, but was confused because this problem has the ladder moving down, but none have one with something going up. Thanks in advance!

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1 Answer 1

Let $x$ be horizontal and $y$ vertical. I would first express the bug position along the ladder as a function of t: $z=3.5t$. Then you need to find out where this point along the ladder is. The foot of the ladder is at $(1.5t, 0)$, so the top of the ladder is at $(0,what?)$. From these two points you should be able to find the $(x,y)$ coordinates of the bug as a function of time, which will answer all your questions.

Added and corrected: the foot of the ladder is at $(1.5t,0)$ and the top is at $(0,\sqrt{900-2.25t^2})$ The bug is $\frac{3.5t}{30}$ of the way along the ladder, so is at $(1.5t(1-\frac{3.5t}{30}),\frac{3.5t}{30}\sqrt{900-2.25t^2})$

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Done-just lost the last $ –  Ross Millikan Mar 15 '11 at 22:30

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