The equation of the curve for which the cartesian subtangent varies as the reciprocal of the square of the abscissa, is

Question

The equation of the curve for which the cartesian subtangent varies as the reciprocal of the square of the abscissa, is (A) x=c ey3 / 3 k (B) x=c ey2 / 3 k (C) y=c ex3 / 3 k (D) None of these

Tardigrade Admin · Accepted Answer

Given: Cartesian sub-tangent propto (1/ text square of abscissa ) i.e., (y/d y / d x)=(k/x2) or (d y/y)=(x2/k) d x Integrating, log y=(x3/3 k)+ log c or y=c ex3 / 3 k

Q. The equation of the curve for which the cartesian subtangent varies as the reciprocal of the square of the abscissa, is

$x = c e^{y^{3} /3 k}$

$x = c e^{y^{2} /3 k}$

$y = c e^{x^{3} /3 k}$

None of these

Given: Cartesian sub-tangent $\propto \frac{1}{square of abscissa}$
i.e., $\frac{y}{d y / d x} = \frac{k}{x ^{2}}$ or
$\frac{d y}{y} = \frac{x ^{2}}{k} d x$
Integrating, $lo g y = \frac{x ^{3}}{3 k} + lo g c$
or $y = c e^{x^{3} /3 k}$

Q. The equation of the curve for which the cartesian subtangent varies as the reciprocal of the square of the abscissa, is

x=cey3/3k

x=cey2/3k

y=cex3/3k

None of these

Given: Cartesian sub-tangent ∝ square of abscissa 1​ i.e., dy/dxy​=x2k​ or ydy​=kx2​dx Integrating, logy=3kx3​+logc or y=cex3/3k

$x = c e^{y^{3} /3 k}$

$x = c e^{y^{2} /3 k}$

$y = c e^{x^{3} /3 k}$

Given: Cartesian sub-tangent $\propto \frac{1}{square of abscissa}$
i.e., $\frac{y}{d y / d x} = \frac{k}{x ^{2}}$ or
$\frac{d y}{y} = \frac{x ^{2}}{k} d x$
Integrating, $lo g y = \frac{x ^{3}}{3 k} + lo g c$
or $y = c e^{x^{3} /3 k}$