A particle is moving along the curve \(y = \displaystyle\frac{3}{4}x^4 + 3\). The point on the curve at which y-coordinate is changing thrice as fast as the x coordinate, is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\left(1, \displaystyle\frac{15}{4}\right)\)
Both coordinates change with time t. Differentiating the curve with respect to t (rate of change, chain rule):
\( \displaystyle \frac{dy}{dt} = 3x^3\,\frac{dx}{dt} \)
The y-coordinate changes thrice as fast as the x-coordinate:
\( \displaystyle \frac{dy}{dt} = 3\,\frac{dx}{dt} \)
Comparing the two:
\(3x^3 = 3\)
\(x^3 = 1\)
\(x = 1\)
The y-coordinate of the point:
\( \displaystyle y = \frac{3}{4}(1)^4 + 3 = \frac{15}{4} \)
Why the other options are incorrect
Option (1) \(\left(1, -\displaystyle\frac{9}{4}\right)\) — Incorrect. The point is not on the curve: at x = 1 the curve has \(y = \frac{15}{4}\).
Option (2) \((0, 3)\) — Incorrect. At x = 0, \(\frac{dy}{dt} = 0\), so y is not changing at all there.
Option (4) \(\left(-1, \displaystyle\frac{15}{4}\right)\) — Incorrect. At x = −1, \(\frac{dy}{dt} = -3\,\frac{dx}{dt}\): y changes at three times the rate but in the opposite direction (one coordinate increases while the other decreases).