Question

If \(u = \sin\left(\frac{x}{y}\right),\ x = e^t,\ y = t^2\), then \[ t^6 \left(\frac{du}{dt}\right)^2 \div e^{2t}(t - 2)^2 = \]

• A. \(2u\)
• B. \(u^2\)
• C. \(1 - u^2\)
• D. \(\cos u\)

Hint:

For composite functions, apply chain rule carefully. Recognize identities like \(\cos^2 x = 1 - \sin^2 x\) to simplify expressions.

Answer 1

The Correct Option is C

We are given: \[ u = \sin\left(\frac{x}{y}\right),\quad x = e^t,\quad y = t^2 \] So: \[ u = \sin\left(\frac{e^t}{t^2}\right) \] Let \(v = \frac{e^t}{t^2}\), then \(u = \sin(v)\) \[ \frac{du}{dt} = \cos(v) \cdot \frac{dv}{dt} \] Now, \[ v = \frac{e^t}{t^2} \Rightarrow \frac{dv}{dt} = \frac{e^t t^2 - 2t e^t}{t^4} = \frac{e^t(t^2 - 2t)}{t^4} \] So: \[ \frac{du}{dt} = \cos\left(\frac{e^t}{t^2}\right) \cdot \frac{e^t(t - 2)}{t^3} \] Now square and multiply by \(t^6\): \[ t^6 \left(\frac{du}{dt}\right)^2 = t^6 \cdot \cos^2\left(\frac{e^t}{t^2}\right) \cdot \left(\frac{e^t(t - 2)}{t^3}\right)^2 \] \[ = \cos^2\left(\frac{e^t}{t^2}\right) \cdot e^{2t}(t - 2)^2 \] Now divide: \[ \frac{t^6 \left(\frac{du}{dt}\right)^2}{e^{2t}(t - 2)^2} = \cos^2\left(\frac{e^t}{t^2}\right) \] But \(u = \sin\left(\frac{e^t}{t^2}\right)\), so: \[ \cos^2\left(\frac{e^t}{t^2}\right) = 1 - \sin^2\left(\frac{e^t}{t^2}\right) = 1 - u^2 \]