If \( x=16at^{\frac{1}{4}} \) and \( y=8at^{\frac{3}{4}} \) then \( \displaystyle\frac{dy}{dx} \) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{3}{2}\sqrt{t} \)
For parametric equations, \( \frac{dy}{dx}=\frac{dy/dt}{dx/dt} \). Here
\( \displaystyle \frac{dx}{dt}=16a\cdot\frac{1}{4}t^{-3/4} \)
\( \displaystyle =4a\,t^{-3/4} \)
\( \displaystyle \frac{dy}{dt}=8a\cdot\frac{3}{4}t^{-1/4} \)
\( \displaystyle =6a\,t^{-1/4} \)
Dividing, with \( t^{-1/4}\div t^{-3/4}=t^{1/2} \),
\( \displaystyle \frac{dy}{dx}=\frac{6a\,t^{-1/4}}{4a\,t^{-3/4}} \)
\( \displaystyle =\frac{3}{2}\sqrt{t} \)