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Ladder sliding down a wall — related rate of height

A 13 ft ladder leans against a vertical wall, with its base on the ground and its top touching the wall. The base is pulled away from the wall at a constant rate of 2 ft/s.

Let xx be the distance from the base of the ladder to the wall, and let yy be the height of the top of the ladder on the wall. Since the ladder has fixed length, xx and yy satisfy x2+y2=132x^2 + y^2 = 13^2. The base slides away from the wall at dxdt=2\frac{dx}{dt} = 2 ft/s. At the instant when x=5x = 5 ft, how fast is the height yy of the top of the ladder decreasing (that is, find ∣dydt∣\left|\frac{dy}{dt}\right| at that moment, in ft/s)?

Hint

Differentiate both sides of x2+y2=169x^2 + y^2 = 169 with respect to tt using the chain rule, then plug in the known values of xx, yy, and dxdt\frac{dx}{dt} to solve for dydt\frac{dy}{dt}. First find yy from the Pythagorean relation when x=5x=5.

Step-by-step solution

Since x2+y2=169x^2+y^2=169, when x=5x=5 we get y2=169−25=144y^2 = 169 - 25 = 144, so y=12y = 12 ft. This uses the fixed ladder length as the constraint equation relating xx and yy at every instant.

Differentiating x2+y2=169x^2+y^2=169 with respect to tt gives 2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0, so dydt=−xydxdt\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}. Substituting x=5x=5, y=12y=12, and dxdt=2\frac{dx}{dt}=2 gives dydt=−512(2)=−56\frac{dy}{dt} = -\frac{5}{12}(2) = -\frac{5}{6} ft/s. The negative sign indicates yy is decreasing, so the top of the ladder drops at a rate of 56\frac{5}{6} ft/s.

Ladder sliding down a wall — related rate of height — solution graph

Direction of height change as ladder slides

Hint
Step-by-step solution

The height is decreasing. Since the ladder length is fixed, as the base distance x grows, the height y must shrink to satisfy x squared plus y squared equals one hundred sixty-nine. This matches the negative sign found for dy/dt in part a, confirming the top of the ladder moves downward as the base moves outward.

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