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.
-$4 x ^ { 2 } y - \pi \cos y = 5 \pi$, slope at $( 1 , \pi )$
A) 0
B) $- 2 \pi$
C) $\pi$
D) $- \frac { \pi } { 2 }$

Question

Quizplus · Accepted Answer

To find the slope at a given point on the curve, we need to differentiate the equation implicitly with respect to $x$, since the equation is not explicitly solved for $y$. Differentiating both sides of the equation $4x^2y - \pi \cos(y) = 5\pi$ with respect to $x$, we get:$$8xy + 4x^2 \frac{dy}{dx} - \pi(-\sin(y))\frac{dy}{dx} = 0$$At the point $(1, \pi)$, we substitute $x = 1$ and $y = \pi$ into the differentiated equation:$$8(1)(\pi) + 4(1)^2 \frac{dy}{dx} - \pi(-\sin(\pi))\frac{dy}{dx} = 0$$Since $\sin(\pi) = 0$, the equation simplifies to:$$8\pi + 4\frac{dy}{dx} = 0$$Solving for $\frac{dy}{dx}$, we get:$$\frac{dy}{dx} = -2\pi$$Therefore, the slope of the curve at the point $(1, \pi)$ is $-2\pi$.

At the Given Point, Find the Slope of the Curve 4x2y−πcos⁡y=5π4 x ^ { 2 } y - \pi \cos y = 5 \pi4x2y−πcosy=5π

At the Given Point, Find the Slope of the Curve $4 x ^ { 2 } y - \pi \cos y = 5 \pi$