A particle moves along a curve \( 3y^2=6x^3+8 \) such that its y-coordinate is changing at the rate of 9 cm/sec at point (1,2). Then the rate of a change of x-coordinate at the same point is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6 cm/sec
Differentiating the equation of the curve with respect to time t (related rates):
\( \displaystyle 6y\frac{dy}{dt}=18x^2\frac{dx}{dt} \)
At (1, 2) with \( \frac{dy}{dt}=9 \):
\( \displaystyle 6(2)(9)=18(1)^2\frac{dx}{dt} \)
\( \displaystyle 108=18\frac{dx}{dt}\ \Rightarrow\ \frac{dx}{dt}=6 \)
So the x-coordinate changes at 6 cm/sec.
Why the other options are incorrect
Option (1) \( \displaystyle\frac{9}{7} \) cm/sec — Incorrect. The rate is 6 cm/sec, not 9/7 cm/sec.
Option (2) 3 cm/sec — Incorrect. 3 cm/sec leaves out the factor 2 from \( \frac{d}{dt}(y^2)=2y\frac{dy}{dt} \): \( 3y\frac{dy}{dt}=18x^2\frac{dx}{dt} \) gives 54 = 18 dx/dt.
Option (3) \( \displaystyle\frac{3}{4} \) cm/sec — Incorrect. The rate is 6 cm/sec, not 3/4 cm/sec.