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October 1, 2014

October 1, 2014

Posted by **Preston** on Monday, November 1, 2010 at 5:17pm.

- Calc. -
**Reiny**, Monday, November 1, 2010 at 6:39pmThis is the classic problem that almost every text in Calculus uses to introduce "rate of change"

Make a diagram

Let the ladder by y ft above the ground, and

x ft away from the wall

so you have a right angled triangle

x^2 + y^2 = 25

2x dx/dt + 2y dy/dt = 0 (#1)

given : dx/dt = 2 ft/s

find : dy/dt when x = 7

when x = 7

49 + y^2 = 625

y = √576 = 24

back in #1

2(7)(2) + 2(24)dy/dt = 0

dy/dt = -28/48 = -7/12 ft/s

the negative indicates that the value of y is decreasing, or in other words, the ladder is dropping along the wall

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