Question

I have this calculus problem that I’m having trouble with. I’m not allowed to use the power rule. I need to use the formula s’(t) = (s(t+h) - s(t))/h. Please help

Answer 1

Step-by-step explanation:

Know that

The derivative of position with respect of t is velocity.

Or if you want to say

$\frac{d}{dt} (s(t) = v$

Or in the form of the definition of the derivative

$s'(t) = \frac{6 (t + h) {}^{3} + 2(t + h) + 9 - (6 {t}^{3} + 2t + 9) }{h}$

$s'(t) = \frac{6 {t}^{3} + 18t {}^{2}h + 18h {}^{2} t + 6 {h}^{3} + 2t + 2h + 9 - (6 {t}^{3} + 2t + 9)}{h}$

$s'(t) = \frac{18 {t}^{2}h + 18h {}^{2}t + 6h {}^{3} + 2h }{h}$

$s'(t) = 18 {t}^{2} + 18th + 6h {}^{2} + 2$

Let h be a small number, so we can cancel out small things.

$s'(t) = 18 {t}^{2} + 2$

We know the derivative of position is velocity

$v = 18 {t}^{2} + 2$

So the velocity at time t is

$v = 18 {t}^{2} + 2$

Plug in 3 for t to find the velocity

$v(3) =18(3) {}^{2} + 2$

$v(3) = 164$

So the velocity at t=3, is 164.

Next, the derivative of velocity is acceleration so

$v'(t) = \frac{18(t + h) {}^{2} + 2 - (18 {t}^{2} + 2)}{h}$

$v'(t) = \frac{18 {t}^{2} + 36ht + 18 {h}^{2} + 2(- 18 {t}^{2} + 2 ) }{h}$

$v'(t) = \frac{36ht + 18 {h}^{2} }{h}$

Divide by h.

$v'(t) = 36t + 18h$

$v'(t) = 36t$

The derivative of velocity is acceleration so

$a = 36t$

The acceleration in time t is

$36t$

To find acceleration, plug in 3 for t.

$a(3) = 36( 3) = 108$