Question

If $z = 3x^2 y^3$, $x = t^4$ and $y = t^2$, then using chain rule $\frac{dz}{dt}$ is given by}

• A. $42 t^{13}$
• B. $52 t^{13}$
• C. $32 t^{13}$
• D. 0

Hint:

For single-variable parameterizations, direct substitution of $x(t)$ and $y(t)$ into $z$ is often faster and less prone to calculation errors than executing the full multi-variable chain rule.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The chain rule for a function of several variables allows us to compute the derivative of a composite function with respect to a single independent parameter.
Key Formula or Approach:
For $z = f(x, y)$ where $x = x(t)$ and $y = y(t)$:
\[ \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \]

Step 2: Detailed Explanation:
Let us first compute the individual derivatives:
- From $z = 3x^2 y^3$:
\[ \frac{\partial z}{\partial x} = 6xy^3 \]
\[ \frac{\partial z}{\partial y} = 9x^2y^2 \]
- From $x = t^4$:
\[ \frac{dx}{dt} = 4t^3 \]
- From $y = t^2$:
\[ \frac{dy}{dt} = 2t \]
Now, substitute these derivatives into the chain rule formula:
\[ \frac{dz}{dt} = (6xy^3) \cdot (4t^3) + (9x^2y^2) \cdot (2t) \]
Substitute $x = t^4$ and $y = t^2$ into the equation:
\[ \frac{dz}{dt} = [6(t^4)(t^2)^3] \cdot (4t^3) + [9(t^4)^2(t^2)^2] \cdot (2t) \]
Simplify the powers of $t$:
\[ \frac{dz}{dt} = [6t^{10}] \cdot (4t^3) + [9t^{12}] \cdot (2t) \]
\[ \frac{dz}{dt} = 24t^{13} + 18t^{13} = 42t^{13} \]
(Alternative verification: Substitute $x$ and $y$ directly into $z$ first: $z = 3(t^4)^2(t^2)^3 = 3t^8 \cdot t^6 = 3t^{14}$. Then $\frac{dz}{dt} = 3 \cdot 14t^{13} = 42t^{13}$.)
Both methods yield $42 t^{13}$.

Step 3: Final Answer
The correct option is (A).