### Math Notes

Subjects

#### Differential Calculus Solutions

##### Topics || Problems

Find the velocity of a point whose position is given by the equation $$s = 14t^3 + 5t -1$$, where t = 2.

Solution:

$$\frac{ds}{dt} = velocity$$

$$s = 14t^3 + 5t -1$$

$$\frac{ds}{dt} = 42t^2 +5$$

At t = 2

$$\frac{ds}{dt} = 42(2)^2 +5$$

$$\frac{ds}{dt} = 173$$