### Math Notes

Subjects

#### Differential Calculus Solutions

##### Topics || Problems

If $$s$$ is measured in feet and $$t$$ in seconds, what is the rate at which $$s$$ is changing at the end of 2 seconds when $$s = \frac{1}{1-t^2}$$.

$$s = \frac{1}{1-t^2}$$

$$\frac{ds}{dt} = \frac{0 -1 (1-t^2)^2}{(1-t^2)}$$

$$\frac{ds}{dt} = v = \frac{-(1-t^2)^2}{1-t^2}$$

At $$t = 2 ~sec$$

$$v = \frac{-(1-4)^2}{1-4}$$

$$v = \frac{9}{4} ~fps$$