Differential Calculus Solutions
Topics || Problems
A 5m long ladder leans against a vertical wall 4m high. If the lower end is sliding at 1 m/sec, how fast is the tip of the ladder moving?
Solution
Solution
By phytagorean formula: \(5^2 = y^2 + x^2 \)
At \(y = 4\) the value of \(x\) is \(3\).
\(0 = 2y\frac{dy}{dt} + 2x\frac{dx}{dt}\)
At \(y = 4\), \(x = 3\), \(\frac{dx}{dt} = 1\)
\(0 = 2(4)\frac{dy}{dt} + 2 (3)(1)\)
\(\frac{dy}{dt} = -\frac{3}{4} \text{m/s}\). The negative sign indicates a downward movement.
Thus the tip of the ladder moves at \(\frac{3}{4} \text{m/s}\) downwards.