Q. The equation of the curve in which the portion of $y$-axis cut off between the origin and the tangent varies as the cube of the abscissa of the point of contact is
A) $y = \frac{k\ x^{3}}{3} + c\ x$
B) $y = - \frac{k\ x^{2}}{2} + c$
C) $y = - \frac{k\ x^{3}}{2} + c\ x$
D) $y = \frac{k\ x^{3}}{3} + \frac{c\ x^{2}}{2}$
$(k$ is constant of proportionality)
(where $c$ is arbitrary constant)
Explanation
(c)
Equation of tangent at $(x,y)$ is
$Y - y = \frac{dy}{dx}(X - x)$
For $y$-axis $X = 0.$
Then, $Y = y - x\frac{dy}{dx}$
Given, $\left( \ y - x\frac{dy}{dx}\ \right) \propto \ x^{3}$
$\Rightarrow \ y - x\frac{dy}{dx} = kx^{3}$
$\Rightarrow \ \frac{dy}{dx} - \frac{y}{x} = - kx^{2}$
IF $= e^{\int_{}^{}{1/x\ dx}} = e^{- In\ x} = e^{In(1/x)} = \frac{1}{x}$
Then, solution is
$y\ \left( \frac{1}{x} \right) = \int_{}^{}{\frac{- kx^{2}}{x}dx}$
$\Rightarrow \frac{y}{x} = - \frac{kx^{2}}{2} + c$
or $y = - \frac{kx^{3}}{2} + cx$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.