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#1 Re: Help Me ! » Related rates problem. Please help! » 2008-10-19 09:55:29
Ok, I think I understand your concept, but are you saying that I need to take the derivative of √(225-(10-t/4)^2) as it is? I cant just plug in 12 for t and solve?
#2 Help Me ! » Related rates problem. Please help! » 2008-10-19 08:03:33
- ashleyrc
- Replies: 3
A 15 foot ladder is resting against the wall. The bottom is initially 10 feet away from the wall and is being pushed towards the wall at a rate of ¼ ft/sec. how fast is the top of the ladder moving up the wall 12 seconds after we started pushing?
A diagram of a triangle was provided, showing that the hypotenuse is 15 ft, and the base is moving towards the wall at a rate of -1/4 (x^t = -1/4)
I know that distance is equal to (rate x time) ¼ x 12 = 3, which means that 10-3 = 7. 7 is the new distance for the base after 12 seconds.
How do I find the rate of change of the ladder moving up the wall? 7 = y^t (y)?
#3 Re: Help Me ! » derivative problem (down to simplifying?) » 2008-10-06 19:01:33
sorry, not when x = 3/. i meant to say when x=3
#4 Help Me ! » derivative problem (down to simplifying?) » 2008-10-06 18:38:01
- ashleyrc
- Replies: 2
find equation of the tangent line to the graph of the function f(x)=square root of x^2+16, when x = 3/
what i did: think i'm supposed to just take the derivative, which would equal f(x) = x + 4 ; f'(x) = 1? does that mean limit does not exist?
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