Question 1044712: CALCULUS(MAXIMA AND MINIMA): A 4m ladder is leaning against a vertical wall with its foot on the same horizontal plane as the base of the wall.if the lower end of the ladder is moving away from the wall horizontally at 1.3m/sec,how fast is the top of the ladder is descending when the lower end is 2m from the wall?
*show the illustration
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! We want dx/dt where x is the distance of the top of the ladder from the ground
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We have dy/dt = 1.3 m/sec where y is the distance between the ladder and the wall
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Using the Pythagorean Theorem, we know that
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4^2 = x^2 + y^2
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we differentiate the above equation with respect to time (t), then
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0 = 2x(dx/dt) + 2y(dy/dt)
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We want dx/dt, so we solve for it
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(dx/dt) = (-2y / 2x)(dy/dt) = (-y/x)(dy/dt)
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The following is given in the problem, x=2 and dy/dt = 1.3 m/sec
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4^2 = 2^2 +y^2
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y = square root(12) = 2*square root(3)
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(dx/dt) = ((-2*square root(3) / 2) * (1.3) = −2.2517 approx -2.25 m/sec
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The top of the ladder is moving towards the ground at 2.25 m/sec
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