The equation of the tangent to the curve 6y = | vedclass

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

Question

(B) The given curve is $6y = 7 - x^3$. To find the slope of the tangent, we differentiate the equation with respect to $x$: $6 \frac{dy}{dx} = -3x^2$

Vedclass · Accepted Answer

$x + 2y = 3$

Vedclass · Answer

(B) The given curve is $6y = 7 - x^3$. To find the slope of the tangent, we differentiate the equation with respect to $x$: $6 \frac{dy}{dx} = -3x^2$ $\frac{dy}{dx} = -\frac{3x^2}{6} = -\frac{x^2}{2}$ At the point $(1, 1)$, the slope $m$ is: $m = -\frac{(1)^2}{2} = -\frac{1}{2}$ The equation of the tangent line passing through $(x_1, y_1) = (1, 1)$ with slope $m = -\frac{1}{2}$ is given by $y - y_1 = m(x - x_1)$: $y - 1 = -\frac{1}{2}(x - 1)$ $2(y - 1) = -(x - 1)$ $2y - 2 = -x + 1$ $x + 2y = 3$

The equation of the tangent to the curve $6y = 7 - x^3$ at the point $(1, 1)$ is:

Similar Questions

Find the equation of the normal to the curve $y = x^3$ at the point $P(1, 1)$.

The length of the subtangent at any point $(x_1, y_1)$ on the curve $y=5^x$ is

The equation of a straight line passing through the point $(3, 6)$ and cutting the curve $y = \sqrt{x}$ orthogonally is

If the normal to the curve ${y^2} = 5x - 1$ at the point $(1, -2)$ is of the form $ax - 5y + b = 0$,then $a$ and $b$ are:

If the equation of the normal to the curve $y = \frac{x-a}{(x+b)(x-2)}$ at the point $(1, -3)$ is $x - 4y = 13$,then the value of $a+b$ is equal to $.......$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module