Math Is Fun Forum

ashleyrc

Member

Registered: 2008-10-06

Posts: 3

Related rates problem. Please help!

“ 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)?

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Related rates problem. Please help!

Use the Pythagorean theorem. C is constant (15 feet, the length of the ladder).

Let A be the distance between the wall and the ladders base.

Let B be the height at which the ladder rests against the wall.

B(t) = √(15^2-(10-t/4)^2) = √(225-(10-t/4)^2)

taking the derivative of B(t) with respect to t, then evaluating at t=12, will give you your answer.

Last edited by All_Is_Number (2008-10-19 08:46:22)

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ashleyrc

Member

Registered: 2008-10-06

Posts: 3

Re: Related rates problem. Please help!

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 can’t just plug in 12 for t and solve?

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Related rates problem. Please help!

Evaluating the expression at t=12 without taking the derivative first will only give you the height at which the ladder rests against the wall after 12 seconds, not the rate of change for that height.

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You can shear a sheep many times but skin him only once.