Question

suppose y = √(2x + 1), where x and y are functions of t. (a) if dx/dt = 9, find dy/dt when x = 4. dy/dt = (b) if dy/dt = 3, find dx/dt when x = 40. dx/dt =

Explanation:

Step1: Differentiate y with respect to x

Using the chain - rule, if $y=\sqrt{2x + 1}=(2x + 1)^{\frac{1}{2}}$, then $\frac{dy}{dx}=\frac{1}{2}(2x + 1)^{-\frac{1}{2}}\times2=\frac{1}{\sqrt{2x + 1}}$.

Step2: Use the chain - rule $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$

(a)

When $x = 4$, $\frac{dy}{dx}=\frac{1}{\sqrt{2\times4+1}}=\frac{1}{\sqrt{9}}=\frac{1}{3}$. Given $\frac{dx}{dt}=9$, then $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}=\frac{1}{3}\times9 = 3$.

(b)

When $x = 40$, $\frac{dy}{dx}=\frac{1}{\sqrt{2\times40+1}}=\frac{1}{\sqrt{81}}=\frac{1}{9}$. Given $\frac{dy}{dt}=3$, from $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$, we can solve for $\frac{dx}{dt}$. So $\frac{dx}{dt}=\frac{\frac{dy}{dt}}{\frac{dy}{dx}}$. Substituting the values, we get $\frac{dx}{dt}=3\div\frac{1}{9}=27$.

Answer:

(a) $\frac{dy}{dt}=3$
(b) $\frac{dx}{dt}=27$