x = 6.0y = 8.0010 ft

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x (ft)
6.0
y = sqrt(100 - x^2)
8.000
dy/dt (ft/s)
-1.5000
Relation
x^2 + y^2 = 10^2 = 100
Differentiate the relation with respect to t
2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0
Given rate
\frac{dx}{dt} = +2 \text{ ft/s}
Solve for the unknown rate
\frac{dy}{dt} = -\frac{2x}{2y}\,\frac{dx}{dt} = -\frac{x}{y}\,\frac{dx}{dt}
Substitute the current state
\frac{dy}{dt} = -\frac{2(6)(2)}{2(8)} = -1.5
Ladder base x = 6.0 feet, height y = 8.00 feet, dy/dt = -1.500 feet per second.

Differentiate the relation before you substitute. The relation holds for every instant, so its derivative in t links the rates. If you plug the current numbers in first, x becomes a fixed constant, its derivative is zero, and the equation can no longer solve for how fast the top slides down the wall.

The drag stops at x = 9. Since dy/dt = -2x / sqrt(100 - x^2) and the height y = sqrt(100 - x^2) heads to 0 as x approaches 10, the speed |dy/dt| grows without bound near the wall. A ladder problem that asks for the rate at the instant the ladder is flat has no finite answer, an AP favorite.

Related Rates Scene: Sliding Ladder and Cone Tank from CalcLearn, free AP Calculus study tools.