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Hunt for the tangent

grgrsanjay

New member
Jan 28, 2012
21
The curve $\displaystyle y-e^{(xy)} + x=0 $ has a vertical tangent at which point??

I started to differentiate it, then equating dy/dx to 0, then how should i proceed??
 
Last edited:

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
The curve $\displaystyle y-e^{(xy)} + x=0 $ has a vertical tangent at which point??
I started to differentiate it, then equating dy/dx to 0, then how should i proceed??
Find where $y'$ is not defined. It is not where it is zero.
 

grgrsanjay

New member
Jan 28, 2012
21
Sorry....i did the same,typed wrongly

i equated the denominator to zero.i got $ xe^{xy} = 1 $

then what?
----------------------------------------------

Yea,yea got it dy/dx=0, so the equation is of the form x = a
y=0

substituting it at this $ xe^{xy} = 1 $,we getx=1

So, point is (1,0)

hmm.....i stumbled to a easy question :(
 
Last edited:

soroban

Well-known member
Feb 2, 2012
409
Hello, grgrsanjay!

I had an idea . . . then hit a wall.
Maybe someone can carry on ?


The curve .$\displaystyle y-e^{xy} + x=0 $ has a vertical tangent at which point?

Differentiate implicitly: .$y' - e^{xy}(xy' + y) + 1 \;=\;0 \quad\Rightarrow\quad y' - xy'e^{xy} - ye^{xy} + 1 \;=\;0 $

. . . . . $ y' - xy'e^{xy} \;=\; ye^{xy} - 1 \quad\Rightarrow\quad y'(1-xe^{xy}) \;=\; ye^{xy}-1 $

. . . . . $y' \;=\;\dfrac{ye^{xy} - 1}{1 - xe^{xy}} $


The curve has a vertical tangent where the denominator equals zero:

. . . $ 1 - xe^{xy} \:=\:0 \quad\Rightarrow\quad e^{xy} \:=\:\dfrac{1}{x} $


Substitute into the original equation (?)

. . . $ y - \frac{1}{x} + x \:=\:0 \quad\Rightarrow\quad y \:=\:\dfrac{1-x^2}{x}$


Does this help?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,189
Hello, grgrsanjay!

I had an idea . . . then hit a wall.
Maybe someone can carry on ?



Differentiate implicitly: .$y' - e^{xy}(xy' + y) + 1 \;=\;0 \quad\Rightarrow\quad y' - xy'e^{xy} - ye^{xy} + 1 \;=\;0 $

. . . . . $ y' - xy'e^{xy} \;=\; ye^{xy} - 1 \quad\Rightarrow\quad y'(1-xe^{xy}) \;=\; ye^{xy}-1 $

. . . . . $y' \;=\;\dfrac{ye^{xy} - 1}{1 - xe^{xy}} $


The curve has a vertical tangent where the denominator equals zero:

. . . $ 1 - xe^{xy} \:=\:0 \quad\Rightarrow\quad e^{xy} \:=\:\dfrac{1}{x} $


Substitute into the original equation (?)

. . . $ y - \frac{1}{x} + x \:=\:0 \quad\Rightarrow\quad y \:=\:\dfrac{1-x^2}{x}$


Does this help?
Hmm. $xy=\ln(1/x)$, so $y=\ln(1/x)/x$. Therefore, solve the following simultaneously:

\begin{align*}
y&=\frac{1-x^{2}}{x}\\
y&=\frac{\ln(1/x)}{x}.
\end{align*}
Hence
$$\frac{1-x^{2}}{x}=\frac{\ln(1/x)}{x}\implies 1-x^{2}=\ln(1/x)\implies e^{1-x^{2}}=1/x\implies e^{x^{2}-1}=x.$$
There look to be two solutions, but I think you might have to find one of them numerically.
 

grgrsanjay

New member
Jan 28, 2012
21
Thanks for your help!

I Understood it :)