Question

Using chain rule, the derivative of $\omega = x^2 y - y^2$ with respect to t along the path $x = \sin t$ and $y = e^t$ at $t = 0$ is given by

• A. 0
• B. -1
• C. 2
• D. -2

Hint:

Always find the values of the intermediate variables ($x$ and $y$) at the specified parameter value ($t = 0$) first. This simplifies the numerical evaluation of the partial derivatives immediately.

Answer 1

The Correct Option is D

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

Step 2: Detailed Explanation:
Given the function $\omega = x^2 y - y^2$:
First, calculate the partial derivatives of $\omega$: \[ \frac{\partial\omega}{\partial x} = 2xy \] \[ \frac{\partial\omega}{\partial y} = x^2 - 2y \]
Next, calculate the derivatives of $x$ and $y$ with respect to $t$: \[ x = \sin t \implies \frac{dx}{dt} = \cos t \] \[ y = e^t \implies \frac{dy}{dt} = e^t \]
At the given point $t = 0$: \[ x(0) = \sin(0) = 0 \] \[ y(0) = e^0 = 1 \]
Now, evaluate the partial derivatives at $t = 0$ (where $x = 0$ and $y = 1$): \[ \frac{\partial\omega}{\partial x} = 2(0)(1) = 0 \] \[ \frac{\partial\omega}{\partial y} = 0^2 - 2(1) = -2 \]
And evaluate the parametric derivatives at $t = 0$: \[ \frac{dx}{dt} = \cos(0) = 1 \] \[ \frac{dy}{dt} = e^0 = 1 \]
Substitute these values into the total derivative formula: \[ \frac{d\omega}{dt} = (0)(1) + (-2)(1) = -2 \]

Step 3: Final Answer:
The derivative of $\omega$ with respect to $t$ at $t = 0$ is $-2$.