Question 1160398
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x is the distance the foot of the 17-foot ladder is from the wall when the height the ladder reaches on the wall is y:<br>
{{{x = sqrt(17^2-y^2) = (289-y^2)^(1/2)}}}<br>
Differentiate to find the rate of change in x for the given rate of change in y.<br>
Square the given equation to avoid differentiating the square root function.<br>
{{{x^2 = 289-y^2}}}
{{{2x*(dx/dt) = -2y*(dy/dt)}}}<br>
dy/dt is given as -4 (ft/sec); when y=14, x is {{{sqrt(289-196) = sqrt(93)}}}.<br>
{{{2*sqrt(93)(dx/dt) = -2(14)(-4)}}}
{{{dx/dt = 112/(2*sqrt(93)) = 56/sqrt(93)}}}<br>
When the top of the ladder is 14 feet above the ground, the foot is moving away from the wall at a rate of 56/sqrt(93) ft/sec, or about 5.8 ft/sec.<br>