Given $y = (2x + 1)^{1/2}$, where $x$ and $y$ are functions of $t$, we can find $\frac{dy}{dt}$ by differentiating both sides of the equation with respect to $t$.

Step 1: Differentiate $y$ with respect to $t$

Using the chain rule: $\frac{dy}{dt} = \frac{d}{dt} \left( (2x + 1)^{1/2} \right)$ Let $u = 2x + 1$, so $y = u^{1/2}$. Then, $\frac{dy}{dt} = \frac{1}{2}(2x + 1)^{-1/2} \cdot \frac{d(2x + 1)}{dt}$ Now, differentiate $2x + 1$ with respect to $t$: $\frac{d(2x + 1)}{dt} = 2 \frac{dx}{dt}$ Thus, $\frac{dy}{dt} = \frac{1}{2}(2x + 1)^{-1/2} \cdot 2 \frac{dx}{dt}$ Simplifying, $\frac{dy}{dt} = \frac{\frac{dx}{dt}}{(2x + 1)^{1/2}}$

(a) Find $\frac{dy}{dt}$ when $x = 4$ and $\frac{dx}{dt} = 3$

Substitute $x = 4$ and $\frac{dx}{dt} = 3$ into the formula for $\frac{dy}{dt}$: $\frac{dy}{dt} = \frac{3}{\sqrt{2 \cdot 4 + 1}} = \frac{3}{\sqrt{8 + 1}} = \frac{3}{\sqrt{9}} = \frac{3}{3} = 1$ So, when $x = 4$, $\frac{dy}{dt} = 1$.

(b) Find $\frac{dx}{dt}$ when $x = 12$ and $\frac{dy}{dt} = 5$

Rearrange the formula for $\frac{dy}{dt}$ to solve for $\frac{dx}{dt}$: $\frac{dy}{dt} = \frac{\frac{dx}{dt}}{\sqrt{2x + 1}}$ $\frac{dx}{dt} = \frac{dy}{dt} \cdot \sqrt{2x + 1}$ Substitute $x = 12$ and $\frac{dy}{dt} = 5$: $\frac{dx}{dt} = 5 \cdot \sqrt{2 \cdot 12 + 1} = 5 \cdot \sqrt{24 + 1} = 5 \cdot \sqrt{25} = 5 \cdot 5 = 25$ So, when $x = 12$, $\frac{dx}{dt} = 25$.

Summary:

• (a) $\frac{dy}{dt} = 1$ when $x = 4$ and $\frac{dx}{dt} = 3$.
• (b) $\frac{dx}{dt} = 25$ when $x = 12$ and $\frac{dy}{dt} = 5$.

Would you like further details on any steps?

---

1. How does the chain rule apply in problems involving multiple variables?
2. What other scenarios would involve differentiating functions of multiple variables with respect to time?
3. How can we interpret $\frac{dy}{dt}$ and $\frac{dx}{dt}$ in real-world contexts?
4. What would change if $y$ were a function of multiple independent variables?
5. How would these results be affected if $dx/dt$ and $dy/dt$ were functions instead of constants?

Tip: Always check if variables are functions of other variables when differentiating implicitly.