Answer to Question #284623 in Calculus for poochie

Question

Use the Chain Rule to determine an equation for the acceleration when 𝑎 = 𝑑𝑣/ 𝑑𝑡

When 𝑣 = (2𝑡² + 3)⁴

and

When 𝑣 = ln(4𝑡³ − 1)

Answer 1

1)When $v = (2t^2+3)^4 = f(g(t))$, where $f(t) = t^4, g(t) = 2t^2 + 3$ .

Since $\frac{df}{dt} = 4t^3$ and $\frac{dg}{dt} = 4t$ , by the chain rule, we obtain that:

$a = \frac{dv}{dt} = \frac{df}{dg} \frac{dg}{dt} = 4(2t^2+3)^3 \cdot 4t = 16t(2t^2+3)^3$

2)When $v = ln(4t^3-1) = f(g(t))$, where $f(t) =ln( 4t^3-1), g(t) = (4t^3 - 1)$ .

Since $\frac{df}{dt} =\frac{1}{4t^3-1}$ and $\frac{dg}{dt} =12t^2$ , by the chain rule, we obtain that:

$a = \frac{dv}{dt} = \frac{df}{dg} \frac{dg}{dt} = \frac{1}{4t^3-1}\cdot 12t^2 = \frac{12t^2}{4t^3-1}$