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

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Question

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

Vedclass · Accepted Answer

$x + 2y = 3$

Vedclass · Answer

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

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