Find equations of the tangent line and normal line to

hvacwk 2021-12-19 Answered
Find equations of the tangent line and normal line to the given curve at the specified point. \(\displaystyle{y}={2}{x}{e}^{{{x}}},\ {\left({0},\ {0}\right)}\)

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Expert Answer

Cleveland Walters
Answered 2021-12-20 Author has 2423 answers
Step 1
\(\displaystyle{f{{\left({x}\right)}}}={2}{x}{e}^{{{x}}}\)
Differentiate
\(\displaystyle{f}'{\left({x}\right)}={\frac{{{d}{\left({2}{x}{e}^{{{x}}}\right)}}}{{{\left.{d}{x}\right.}}}}\)
\(\displaystyle{f}'{\left({x}\right)}={2}{x}\times{\frac{{{d}{\left({e}^{{{x}}}\right)}}}{{{\left.{d}{x}\right.}}}}+{e}^{{{x}}}\times{\frac{{{d}{\left({2}{x}\right)}}}{{{\left.{d}{x}\right.}}}}\)
\(\displaystyle{2}{x}\times{\frac{{{d}{\left({e}^{{{x}}}\right)}}}{{{\left.{d}{x}\right.}}}}+{e}^{{{x}}}\times{\frac{{{d}{\left({2}{x}\right)}}}{{{\left.{d}{x}\right.}}}}\) Using the product Rule
\(\displaystyle{f}'{\left({x}\right)}={2}{x}\times{\left({e}^{{{x}}}\right)}+{e}^{{{x}}}\times{\left({2}\right)}\)
\(\displaystyle{f}'{\left({x}\right)}={2}{x}\times{e}^{{{x}}}+{2}{e}^{{{x}}}\)
\(\displaystyle{f}'{\left({x}\right)}={e}^{{{x}}}{\left({2}{x}+{2}\right)}\)
Step 2
Slope of the tangent at \(\displaystyle{\left({0},\ {0}\right)}={f}'{\left({0}\right)}={e}^{{{0}}}\times{\left({0}+{2}\right)}={2}\)
Equation of the tangent is
\(\displaystyle{\frac{{{y}-{0}}}{{{x}-{0}}}}={2}\)
\(\displaystyle{y}={2}{x}\)
Step 3
Product of slopes of perpendicular lines is -1
Therefore slope of the normal line is \(\displaystyle-{\frac{{{1}}}{{{2}}}}\)
Equation of the normal line is
\(\displaystyle{\frac{{{y}-{0}}}{{{x}-{0}}}}=-{\frac{{{1}}}{{{2}}}}{x}\)
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Robert Pina
Answered 2021-12-21 Author has 1798 answers
Step 1
The given equation is,
\(\displaystyle{y}={2}{x}{e}^{{{x}}}\)
Step 2
The slope is obtained as follows.
\(\displaystyle{y}={2}{x}{e}^{{{x}}}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={2}{\left({e}^{{{x}}}+{x}{e}^{{{x}}}\right)}\)
\(\displaystyle{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\right)}_{{{x}={0}}}={2}{\left({e}^{{{0}}}+{0}\right)}={2}\)
\(\displaystyle{m}={2}\)
Step 3
The equation of tangent line is computed as follows.
\(\displaystyle{\left({y}-{y}_{{{1}}}\right)}={m}{\left({x}-{x}_{{{1}}}\right)}\)
\(\displaystyle{\left({y}-{0}\right)}={2}{\left({x}-{0}\right)}\)
\(\displaystyle{y}={2}{x}\)
Step 4
The equation of normal line is computed as follows.
\(\displaystyle{\left({y}-{y}_{{{1}}}\right)}=-{\frac{{{1}}}{{{m}}}}{\left({x}-{x}_{{{1}}}\right)}\)
\(\displaystyle{\left({y}-{0}\right)}=-{\frac{{{1}}}{{{2}}}}{\left({x}-{0}\right)}\)
\(\displaystyle{2}{y}=-{x}\)
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nick1337
Answered 2021-12-27 Author has 9672 answers

Step 1
To obtain slope, we will differentiate the given equation using product rule:
\(\frac{dy}{dx}=\frac{d}{dx}(2xe^{x})\)
\(\frac{dy}{dx}=2x\frac{d(e^{x})}{dx}+e^{x}\times\frac{d(2x)}{dx}\)
\(\frac{dy}{dx}=2xe^{x}+e^{x}\times2\)
\(\frac{dy}{dx}=2xe^{x}+2e^{x}\)
\(\frac{dy}{dx}=e^{x}(2x+2)\)
Now, slope of tangent at \((0,\ 0)=e^{0}(2\times0+2)\)
NSK
So, m=2
Step 2
Now, the equation of tangent line
\(m=\frac{y-y_{1}}{x-x_{1}}\)
\(2=\frac{y-0}{x-0}\)
So, Equation of tangent line is y=2x
Step 3
We know that the slope of normal line is additive inverse of reciprocal of slope of tangent line.
Slope of normal line, \(m=\frac{-1}{2}\)
Now, the equation of normal line
\(m=\frac{y-y_{1}}{x-x_{1}}\)
\(\frac{-1}{2}=\frac{y-0}{x-0}\)
So, Equation of normal line is \(y=\frac{-1}{2}x\)

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