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The equation of a curve is $y=x^2e^{-x}$.

  1. Find the x-coordinate of the stationary points of the curve and determine the nature of these stationary points.
  2. Show that the equation of the normal to the curve at the point where $x=1$ is $e^2x+ey = 1+e^2$.

This is the full question I am having difficulty solving, I simply don't know where to begin. I moved back to my home country because of covid and now I am doing self-studying I don't know how to solve this any help wold be great and much appreciated.

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2 Answers 2

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$y=x^2e^{-x} \implies y'(x)=2xe^{-x}-x^2e^{-x} \implies y'(1)=e^{-1}.$ So the slope of normal at $x=1$ is $m=-1/y'(1)=-e$. So the equation of line having slope $-e$ and passing through the point $(1,e^{-1})$ is $$y-e^{-1}=-e(x-1) \implies ey+e^2 x=1+e^2$$

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  • $\begingroup$ I am confused how did you get this answer? $\endgroup$
    – mary james
    Feb 3, 2021 at 10:46
  • $\begingroup$ @maryjames In $R^2$, if line-1 has slope $m$, and line-2 is perpendicular to line-1, then line-2 has slope $\frac{-1}{m}$. Is this assertion acceptable to you? Assuming so, then the rest of Z Ahmed's answer is nothing more than determining the equation of a line given the slope of the line and a single point on the line. $\endgroup$ Feb 3, 2021 at 12:09
  • $\begingroup$ I am confusedon how you got the gradient in 2 ? could you help me? – $\endgroup$
    – mary james
    Feb 4, 2021 at 10:08
  • $\begingroup$ I don't get it how did you get slope -e? $\endgroup$
    – mary james
    Feb 4, 2021 at 10:14
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    $\begingroup$ $y'(x_0)$ is the slope of tangent at $x=x_0$ and hence slope of normal at $x=x_0$ is $m=-1/y'(x_0),$, here $x_0=1.$ $\endgroup$
    – Z Ahmed
    Feb 4, 2021 at 10:16
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  1. The stationary point of a curve is simply the point where the derivative vanishes. You can see the behavior of these points from this link. https://mathworld.wolfram.com/StationaryPoint.html

The first question asks for the $x$-coordinates of the stationary points. You can start by taking the derivative of the curve. Then, you should equate it to zero and solve the equation to find the roots.

$$\frac{dy}{dx}=2xe^{-x}-x^2e^{-x}=e^{-x}x(2-x)=0\space \Rightarrow \space x=0 \space\text{or}\space x=2$$

At $x=0$ and $x=2$, the derivative is zero which means they are the stationary points.

  1. We should verify the equation for the normal line of the curve at $x=1$. At this point $y$-coordinates of the normal line and the original curve is the same. Since tangent and normal lines to a curve at some particular point are perpendecular to each other, multiplication of their slopes yields $-1$. At $x=1$, the slope of the tangent line is $\frac{1}{e}$. This means the slope of the normal line is $-e$. Below you may see the normal line equation. $$ax+b=y$$ $$x_0=1,\space \space a=-e, \space \space y_0=\frac{1}{e}$$ If we substitute these values into the line equation, we find $b$ to be $e+\frac{1}{e}$. $$-ex+e+\frac{1}{e}=y \space \rightarrow \space e^2+1=e^2x+ey$$
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    $\begingroup$ I'd suggest to separate two steps in (1.) which you enclosed in one line, though named only one of them: 'start by taking a derivative dy/dx=....' then: 'and seek its zeros dy/dx=0 which results in x=...' $\endgroup$
    – CiaPan
    Feb 4, 2021 at 9:04
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    $\begingroup$ I guess you are right. I will edit the answer and make the required explanations. Thanks for the feedback! $\endgroup$ Feb 4, 2021 at 9:34
  • $\begingroup$ I am confusedon how you got the gradient in 2 ? could you help me? $\endgroup$
    – mary james
    Feb 4, 2021 at 10:02
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    $\begingroup$ I noticed a small mistake in my solution so I edited it once again. But I didn't get the gradient part. Since this is a single variable function, the gradient reduces to derivative. $\endgroup$ Feb 4, 2021 at 10:26
  • $\begingroup$ Thank you really appricate it $\endgroup$
    – mary james
    Feb 4, 2021 at 11:09

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