Question

If $x = a \cos^3 t$ and $y = a \sin^3 t$, then the value of $\frac{dy}{dx}$ at $t = \frac{\pi}{4}$ is:

• A. $-1$
• B. $1$
• C. $-\sqrt{3}$
• D. $\frac{1}{\sqrt{3}}$

Hint:

This is the parametric representation of an astroid. Differentiating gives the beautiful, simple result $\frac{dy}{dx} = -\tan t$.

Answer 1

The Correct Option is A

Step 1: Concept
For parametric equations $x = f(t)$ and $y = g(t)$, the derivative is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

Step 2: Meaning
We differentiate both $x$ and $y$ with respect to the parameter $t$, find their ratio, and then substitute $t = \frac{\pi}{4}$.

Step 3: Analysis
Differentiating $x$ with respect to $t$: \[ \frac{dx}{dt} = 3a\cos^2 t (-\sin t) = -3a\cos^2 t \sin t \] Differentiating $y$ with respect to $t$: \[ \frac{dy}{dt} = 3a\sin^2 t (\cos t) = 3a\sin^2 t \cos t \] Now find $\frac{dy}{dx}$: \[ \frac{dy}{dx} = \frac{3a\sin^2 t \cos t}{-3a\cos^2 t \sin t} = -\frac{\sin t}{\cos t} = -\tan t \] At $t = \frac{\pi}{4}$: \[ \frac{dy}{dx} = -\tan\left(\frac{\pi}{4}\right) = -1 \]

Step 4: Conclusion
The derivative of the parametric functions at $t = \frac{\pi}{4}$ is $-1$. Final Answer: (A)