If the slope of the tangent to the curve y = | vedclass

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

Question

(A) Given the curve $y = x^3$. Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = 3x^2$. According to the problem,the slope of the tang

Vedclass · Accepted Answer

$(3, 27)$

Vedclass · Answer

(A) Given the curve $y = x^3$. Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = 3x^2$. According to the problem,the slope of the tangent at a point $(x, y)$ is equal to the $y$-coordinate of that point. So,$\frac{dy}{dx} = y$. Substituting the values,we have $3x^2 = x^3$. This implies $x^3 - 3x^2 = 0$,which factors to $x^2(x - 3) = 0$. Thus,$x = 0$ or $x = 3$. If $x = 0$,then $y = 0^3 = 0$. The point is $(0, 0)$. If $x = 3$,then $y = 3^3 = 27$. The point is $(3, 27)$. Comparing with the given options,$(3, 27)$ is the correct point.

If the slope of the tangent to the curve $y = x^3$ at a point is equal to the $y$-coordinate of that point,then the point is:

Similar Questions

If the normal to the curve $y=f(x)$ at the point $(3,4)$ makes an angle $\left(\frac{3 \pi}{4}\right)^{C}$ with the positive $X$-axis,then $f^{\prime}(3)$ is equal to

The normals to the curve $y=x^{2}-x+1$, drawn at the points with the abscissae $x_{1}=0, x_{2}=-1$ and $x_{3}=5/2$, are:

Find the equations of the tangent and normal to the given curve $y=x^{2}$ at the point $(0,0)$.

Find the condition for the curves $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $xy=c^{2}$ to intersect orthogonally.

The equation of the normal to the curve $2x^{2} + y^{2} = 12$ at the point $(2, 2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module