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.
-$y ^ { 6 } + x ^ { 3 } = y ^ { 2 } + 10 x$, normal at $( 0,1 )$
A) $y = - \frac { 2 } { 5 } x + 1$
B) $y = - \frac { 4 } { 5 } x$
C)$y = \frac { 5 } { 2 } x + 1$
D) $y = \frac { 4 } { 5 } x + 1$

Question

Quizplus · Accepted Answer

To find the slope of the normal line at the point (0,1), we first need to find the slope of the tangent line at that point by differentiating the given equation with respect to $x$. The given equation is $y^6 + x^3 = y^2 + 10x$. Differentiating both sides with respect to $x$, we get:$$6y^5 \frac{dy}{dx} + 3x^2 = 2y \frac{dy}{dx} + 10$$At the point (0,1), we substitute $y = 1$ and $x = 0$ into the differentiated equation:$$6(1)^5 \frac{dy}{dx} + 3(0)^2 = 2(1) \frac{dy}{dx} + 10$$Simplifying, we get:$$6\frac{dy}{dx} = 2\frac{dy}{dx} + 10$$$$4\frac{dy}{dx} = 10$$$$\frac{dy}{dx} = \frac{10}{4} = \frac{5}{2}$$The slope of the tangent line at (0,1) is $\frac{5}{2}$. The slope of the normal line is the negative reciprocal of the tangent slope, which is $-\frac{2}{5}$.The equation of the normal line can be found using 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 the line passes through. Substituting $m = -\frac{2}{5}$ and $(x_1, y_1) = (0, 1)$, we get:$$y - 1 = -\frac{2}{5}(x - 0)$$$$y = -\frac{2}{5}x + 1$$Therefore, the correct answer is $y = -\frac{2}{5}x + 1$, which is option A.

At the Given Point, Find the Slope of the Curve y6+x3=y2+10xy ^ { 6 } + x ^ { 3 } = y ^ { 2 } + 10 xy6+x3=y2+10x

At the Given Point, Find the Slope of the Curve $y ^ { 6 } + x ^ { 3 } = y ^ { 2 } + 10 x$