At the given point, find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve,
as requested.
-$6 x ^ { 2 } y - \pi \cos y = 7 \pi$, tangent at $( 1 , \pi )$
A) $y = - 2 \pi x + 3 \pi$
B) $y = \pi x$
C) $y = - \frac { \pi } { 2 } x + \frac { 3 \pi } { 2 }$
D) $y = - 2 \pi x + \pi$

Question

Quizplus · Accepted Answer

To find the equation of the tangent line at the point (1, π) on the curve $6x^2y - \pi \cos y = 7\pi$, we first need to find the derivative of the given equation implicitly with respect to x to determine the slope of the tangent line at that point. Differentiating both sides of the equation with respect to x gives:$$12xy + 6x^2 \frac{dy}{dx} + \pi \sin y \frac{dy}{dx} = 0$$At the point (1, π), we substitute x = 1 and y = π. Since $\sin(\pi) = 0$, the equation simplifies to:$$12\pi + 6\pi \frac{dy}{dx} = 0$$Solving for $\frac{dy}{dx}$ gives:$$\frac{dy}{dx} = -\frac{12\pi}{6\pi} = -2$$Now, we have the slope of the tangent line at (1, π) which is -2. The equation of the tangent line can be written in point-slope form:$$y - y_1 = m(x - x_1)$$Substituting $m = -2$, $x_1 = 1$, and $y_1 = \pi$, we get:$$y - \pi = -2(x - 1)$$Simplifying, we find:$$y = -2x + 2 + \pi$$$$y = -2x + \pi(1 + 2)$$$$y = -2\pi x + 3\pi$$Therefore, the correct equation of the tangent line at the point (1, π) is $y = -2\pi x + 3\pi$, which matches option A.

At the Given Point, Find the Slope of the Curve 6x2y−πcos⁡y=7π6 x ^ { 2 } y - \pi \cos y = 7 \pi6x2y−πcosy=7π

At the Given Point, Find the Slope of the Curve $6 x ^ { 2 } y - \pi \cos y = 7 \pi$