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.
-$x ^ { 5 } y ^ { 5 } = 32$, tangent at $( 2,1 )$
A) $y = 16 x + 1$
B) $y = 16 x - 1$
C) $y = - \frac { 1 } { 2 } x + 2$
D) $y = \frac { 1 } { 2 } x$

Question

Quizplus · Accepted Answer

To find the slope of the tangent line at the point (2,1), we first need to differentiate the given equation implicitly with respect to $x$. Given $x^5y^5 = 32$, differentiating both sides with respect to $x$ gives:$$5x^4y^5 + 5x^5y^4\frac{dy}{dx} = 0$$Solving for $\frac{dy}{dx}$ (the slope of the tangent line) at the point (2,1):$$5(2)^4(1)^5 + 5(2)^5(1)^4\frac{dy}{dx} = 0$$$$80 + 160\frac{dy}{dx} = 0$$$$\frac{dy}{dx} = -\frac{1}{2}$$The slope of the tangent line at (2,1) is $-\frac{1}{2}$. To find the equation of the tangent line, we use the point-slope form of a line, $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the point on the line:$$y - 1 = -\frac{1}{2}(x - 2)$$$$y - 1 = -\frac{1}{2}x + 1$$$$y = -\frac{1}{2}x + 2$$Therefore, the correct answer is $y = -\frac{1}{2}x + 2$, which matches choice C.

At the Given Point, Find the Slope of the Curve x5y5=32x ^ { 5 } y ^ { 5 } = 32x5y5=32

At the Given Point, Find the Slope of the Curve $x ^ { 5 } y ^ { 5 } = 32$