Find the equation of the normals to the curve | vedclass

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Question

(A) The equation of the given curve is $y=x^{3}+2x+6$. The slope of the tangent to the curve at any point $(x, y)$ is $\frac{dy}{dx} = 3x^{2}+2$. The

Vedclass · Accepted Answer

$x+14y-254=0$ and $x+14y+86=0$

Vedclass · Answer

(A) The equation of the given curve is $y=x^{3}+2x+6$. The slope of the tangent to the curve at any point $(x, y)$ is $\frac{dy}{dx} = 3x^{2}+2$. The slope of the normal at any point $(x, y)$ is $-\frac{1}{\frac{dy}{dx}} = -\frac{1}{3x^{2}+2}$. The equation of the given line is $x+14y+4=0$,which can be written as $y = -\frac{1}{14}x - \frac{4}{14}$. The slope of this line is $-\frac{1}{14}$. Since the normal is parallel to the line,their slopes must be equal: $-\frac{1}{3x^{2}+2} = -\frac{1}{14} \Rightarrow 3x^{2}+2 = 14 \Rightarrow 3x^{2} = 12 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$. For $x=2$,$y = (2)^{3} + 2(2) + 6 = 8 + 4 + 6 = 18$. The point is $(2, 18)$. For $x=-2$,$y = (-2)^{3} + 2(-2) + 6 = -8 - 4 + 6 = -6$. The point is $(-2, -6)$. The equation of the normal at $(2, 18)$ with slope $-\frac{1}{14}$ is: $y - 18 = -\frac{1}{14}(x - 2) \Rightarrow 14y - 252 = -x + 2 \Rightarrow x + 14y - 254 = 0$. The equation of the normal at $(-2, -6)$ with slope $-\frac{1}{14}$ is: $y - (-6) = -\frac{1}{14}(x - (-2)) \Rightarrow y + 6 = -\frac{1}{14}(x + 2) \Rightarrow 14y + 84 = -x - 2 \Rightarrow x + 14y + 86 = 0$. Thus,the equations are $x+14y-254=0$ and $x+14y+86=0$.

Find the equation of the normals to the curve $y=x^{3}+2x+6$ which are parallel to the line $x+14y+4=0$.

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