Question

Given a function \(f(x, y) = \frac{x}{a}e^y + \frac{y}{b}e^x\), where \(x = at\) and \(y = bt\) (a and b are non-zero constants), the value of \(\frac{df}{dt}\) at \(t = 0\) is

• A. -1
• B. 0
• C. 1
• D. 2

Hint:

When applying the chain rule, it's often easier to first find the general expression for \(\frac{df}{dt}\) in terms of \(x\) and \(y\), and only then substitute the specific values of \(x\) and \(y\) corresponding to the given value of \(t\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The function \(f\) is a function of two variables, \(x\) and \(y\), which are themselves functions of a single variable \(t\). To find the total derivative of \(f\) with respect to \(t\), we must use the multivariable chain rule.
Step 2: Key Formula or Approach:
The chain rule for a function \(f(x(t), y(t))\) is given by: \[ \frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} \] Step 3: Detailed Explanation:
1. Find the derivatives of x and y with respect to t: Given \(x = at\) and \(y = bt\). \[ \frac{dx}{dt} = a \] \[ \frac{dy}{dt} = b \] 2. Find the partial derivatives of f with respect to x and y: Given \(f(x, y) = \frac{x}{a}e^y + \frac{y}{b}e^x\). \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( \frac{x}{a}e^y + \frac{y}{b}e^x \right) = \frac{1}{a}e^y + \frac{y}{b}e^x \] \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( \frac{x}{a}e^y + \frac{y}{b}e^x \right) = \frac{x}{a}e^y + \frac{1}{b}e^x \] 3. Apply the chain rule: Substitute the derivatives into the chain rule formula: \[ \frac{df}{dt} = \left( \frac{1}{a}e^y + \frac{y}{b}e^x \right) (a) + \left( \frac{x}{a}e^y + \frac{1}{b}e^x \right) (b) \] \[ \frac{df}{dt} = e^y + \frac{ay}{b}e^x + \frac{bx}{a}e^y + e^x \] 4. Evaluate \(\frac{df}{dt}\) at \(t=0\): First, find the values of \(x\) and \(y\) at \(t=0\). \[ x(0) = a(0) = 0 \] \[ y(0) = b(0) = 0 \] Now substitute \(t=0\), \(x=0\), and \(y=0\) into the expression for \(\frac{df}{dt}\): \[ \frac{df}{dt}\bigg|_{t=0} = e^0 + \frac{a(0)}{b}e^0 + \frac{b(0)}{a}e^0 + e^0 \] \[ \frac{df}{dt}\bigg|_{t=0} = 1 + 0 + 0 + 1 = 2 \] Step 4: Final Answer:
The value of \(\frac{df}{dt}\) at \(t=0\) is 2. Therefore, option (D) is correct.