Need help please!

[sigma]

I have done most of the work for this question and are stuck very near the end of the question and need help.

Find the equation of the normal line to the curve with equation\(\displaystyle \
\L\
xe^y + x\ln (y + e) = 4
\\) at the point with coordinates (2,0).

So here's what I have done...

\(\displaystyle \
\L\
\begin{array}{l}
xe^y + x\ln (y + e) = 4 \\
\to e^y + xe^y (\frac{{dy}}{{dx}}) + \ln (y + e) + x\frac{1}{{y + e}}(\frac{{dy}}{{dx}}) = 0 \\
\to e^0 + 2(e^0 )(m) + \ln (e) + 2 \cdot \frac{1}{e}(m) = 0 \\
\to 2 + 2m + \frac{{2m}}{e} = 0 \\
\to \frac{{2me + 2m}}{e} = - 2 \\
\to 2me + 2m = - 2e??? \\
\end{array}
\\)
and then I get stuck after that. What am I suppose to do with that?

 

[pka]

\(\displaystyle \L
2me + 2m = - 2e\quad \Rightarrow \quad m = \frac{{ - e}}{{e + 1}}\)

 

[soroban]

Hello, sigma!

Find the equation of the normal line to the curve:\(\displaystyle \,xe^y\,+\,x\ln (y + e)\:=\:4\,\) at the point (2,0).

So here's what I have done: \(\displaystyle \,xe^y\,+\,x\ln(y + e)\:=\:4\)

Differentiate implicitly: \(\displaystyle \,e^y\,+\,xe^y\left(\frac{dy}{dx}\right)\,+\,\ln(y\,+\,e)\,+\,x\cdot\frac{1}{y\,+\,e}\left(\frac{dy}{dx}\right)\:=\:0\)

At (2,0), we have: \(\displaystyle \,e^0\,+\,2(e^0 )(m)\,+\,\ln(e)\,+\,2\cdot\frac{1}{e}(m)\:=\:0\)

\(\displaystyle 2\,+\,2m\,+\,\frac{{2m}}{e}\:=\:0\;\;\Rightarrow\;\;\frac{2me\,+\,2m}{e}\:=\:-2\;\;\Rightarrow\;\;2me\,+\,2m\:=\:-2e\)

What am I suppose to do with that? \(\displaystyle \;\Rightarrow\;\;m\:=\:\frac{-e}{e\,+\,1}\)

You've lost sight of your mission . . . Write the equation of the normal.
To write the equation of the normal at (2,0), we'll need the slope of the normal.
We know the normal is perpendicular to the tangent, so we want the slope of the tangent at (2,0).

And that is what you were looking for up there . . . remember?

 

[sigma]

I see now. Don't know why I couldn't before. You got rid of the 2's on both sides because its common. Factored the m out of the LHS then divided by e + 1. Sometimes those are hard to see because your so caught up with everything else. Because its a normal line, wouldn't the answer than be: \(\displaystyle \
\L\
\frac{{ - (e + 1)}}{e}
\\)
which is the inverse of before?

 

[sigma]

Is that right? I just take the inverse of the slope I got which would be \(\displaystyle \
\L\
\frac{{ - (e + 1)}}{e}
\\)?

 

[pka]

If the slope of the tangent is \(\displaystyle m_T = \frac{{ - e}}{{e + 1}}\) then the slope of the normal is \(\displaystyle m_N = \frac{{e + 1}}{e}\).

You see it must be the case that: \(\displaystyle m_T m_N = - 1\).