What is the length of the sub-tangent at any | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of the curve is $y = ax^3$.The length of the sub-tangent at any point $(x_1, y_1)$ is given by the formula $L = |y_1 \cdot \frac{dx}{

Vedclass · Accepted Answer

$x/3$

Vedclass · Answer

(B) The equation of the curve is $y = ax^3$.The length of the sub-tangent at any point $(x_1, y_1)$ is given by the formula $L = |y_1 \cdot \frac{dx}{dy}|$.First,find the derivative of the curve with respect to $x$:$\frac{dy}{dx} = 3ax^2$.Therefore,$\frac{dx}{dy} = \frac{1}{3ax^2}$.At the point $(x_1, y_1)$,the length of the sub-tangent is:$L = |y_1 \cdot \frac{1}{3ax_1^2}|$.Since the point $(x_1, y_1)$ lies on the curve,we have $y_1 = ax_1^3$.Substituting $y_1$ into the formula:$L = |ax_1^3 \cdot \frac{1}{3ax_1^2}| = |\frac{x_1}{3}|$.Thus,the length of the sub-tangent is $x/3$.

What is the length of the sub-tangent at any point on the curve $y = ax^3$?

Similar Questions

If at any point on a curve the subtangent and subnormal are equal,then the length of the normal is equal to

If the normal to the curve $x^{2/3} + y^{2/3} = a^{2/3}$ makes an angle $\phi$ with the $X$-axis,then the equation of that normal is

Find the equation of the normal to the curve $y = \frac{x-7}{(x-2)(x-3)}$ at the point where it cuts the $X$-axis.

If the line $ax + by + c = 0$ is normal to the curve $xy = 1$,then

If the slope of the tangent to the curve $xy + ax + by = 0$ at the point $(1, 1)$ is $2$,then the value of $3a + b$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module