# Thread: stationary pt. of a curve

1. ## stationary pt. of a curve

Help me with this.

Can you tell me what is your value of x.Thanks
i dunno where went wrong..

2. Can't read the text in the figure.

To get the stationary point, find the dy/dx and then set that to zero. The x there will give the stationary point.

3. Originally Posted by maybeline9216
Help me with this. Thanks
Fortunately I have a degree in Egyptology majoring in heiroglyphics. The curve is $\displaystyle y = 3 e^{-x/2} + e^{x/2}$.

Solve $\displaystyle \frac{dy}{dx} = 0$ for x and substitute the value into y:

$\displaystyle 0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}$

$\displaystyle \Rightarrow 0 = -3 + e^x \Rightarrow x = \ln 3$.

Therefore $\displaystyle y = 3 e^{-(\ln 3) /2} + e^{(\ln 3)/2} = 3 = 3 e^{\ln \left(1/\sqrt{3}\right)} + e^{\ln \sqrt{3}} = \frac{3}{\sqrt{3}} + \sqrt{3} = 2 \sqrt{3}$.

4. Originally Posted by ticbol
Can't read the text in the figure.

To get the stationary point, find the dy/dx and then set that to zero. The x there will give the stationary point.
Click the attached file and then click it again to get a better view

5. Originally Posted by maybeline9216
Click the attached file and then click it again to get a better view
Hmmpph .... So in fact only my mathematical skills were required.

6. $\displaystyle 0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}$
can you tell me how to get the value of x from here.

Its because i tried to use the substitution method to get x

For eg. Let x=e^x/2
But i got a different answer.Can you show me another method?

P.S.My dy/dx is correct

7. Originally Posted by mr fantastic
Fortunately I have a degree in Egyptology majoring in heiroglyphics.
Wow! What is that?? lols

8. Originally Posted by maybeline9216
$\displaystyle 0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}$
can you tell me how to get the value of x from here.

Its because i tried to use the substitution method to get x

For eg. Let x=e^x/2
But i got a different answer.Can you show me another method?

P.S.My dy/dx is correct
Originally Posted by Mr Fantastic

[snip]
$\displaystyle 0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}$

Mr F edit: Multiply both sides by $\displaystyle {\color{red}2 e^{x/2}}$:

$\displaystyle \Rightarrow 0 = -3 + e^x \Rightarrow x = \ln 3$.

[snip]
..

9. Thanks!!that's so much easier!

10. $\displaystyle 3 = 3 e^{\ln \left(1/\sqrt{3}\right)} + e^{\ln \sqrt{3}} = \frac{3}{\sqrt{3}} + \sqrt{3} = 2 \sqrt{3}$

Sorry.... i have another qn, how do u get this(see above) ?

11. Originally Posted by maybeline9216
$\displaystyle 3 = 3 e^{\ln \left(1/\sqrt{3}\right)} + e^{\ln \sqrt{3}} = \frac{3}{\sqrt{3}} + \sqrt{3} = 2 \sqrt{3}$

Sorry.... i have another qn, how do u get this(see above) ?
$\displaystyle e^{\ln A} = A \, ....$