### Math Notes

Subjects

#### 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

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.