Checking nature of turning point of parametric equation

[songoku]

Homework Statement

A curve is defined parametrically by
$$x=-(t^2+4)^{\frac{1}{2}} , y=\frac{\ln t}{t}$$

Find the turning point and explain why it is maximum

Relevant Equations

Derivative

Second derivative to check the nature

Sign Diagram

I have found the turning point. I want to ask how to check the nature of the turning point.

My idea is to change the equation into cartesian form then find the second derivative and put the $x$ value of the turning point. If second derivative is positive, then it is minimum and if the second derivative is negative, then it is maximum.

I want to ask whether there is other method to check the nature, maybe directly using the parametric equation (without changing it into cartesian equation).

Thanks

 

[Mark44]

Homework Statement:: A curve is defined parametrically by
$$x=-(t^2+4)^{\frac{1}{2}} , y=\frac{\ln t}{t}$$

Find the turning point and explain why it is maximum
Relevant Equations:: Derivative

Second derivative to check the nature

Sign Diagram

I have found the turning point. I want to ask how to check the nature of the turning point.

My idea is to change the equation into cartesian form then find the second derivative and put the $x$ value of the turning point. If second derivative is positive, then it is minimum and if the second derivative is negative, then it is maximum.

I want to ask whether there is other method to check the nature, maybe directly using the parametric equation (without changing it into cartesian equation).

Thanks

You don't really need to calculate the second derivative. Instead, see what dy/dx does to the left and right of the turning point. Keep in mind, though, that this problem is a little tricky due to the orientation of the parametric path.

 

[songoku]

You don't really need to calculate the second derivative. Instead, see what dy/dx does to the left and right of the turning point. Keep in mind, though, that this problem is a little tricky due to the orientation of the parametric path.

dy/dx in paramateric form or in cartesian form?
Thanks

 

[Mark44]

dy/dx in paramateric form or in cartesian form?
Thanks

I calculated dy/dx as a function of the parameter, t.

 

[songoku]

I calculated dy/dx as a function of the parameter, t.

I get:
$$\frac{dy}{dx}=\frac{(\ln t -1) \sqrt{t^2 + 4}}{t^3}$$

Checking sign diagram for dy/dx for $t=e$, I get negative for left part of $t=e$ and positive for right part of it so the nature is minimum, which contradicts the question

Where is my mistake? Thanks

 

[Mark44]

I get:
$$\frac{dy}{dx}=\frac{(\ln t -1) \sqrt{t^2 + 4}}{t^3}$$

Checking sign diagram for dy/dx for $t=e$, I get negative for left part of $t=e$ and positive for right part of it so the nature is minimum, which contradicts the question

Where is my mistake? Thanks

I get the same. I don't believe you have a mistake. What seems to be happening here is that in your Cartesian graph (x and y axes), as t increases, the points along the Cartesian graph move from right to left. I.e., the part of the x-y graph for t > e has a slope that is positive, while the part of the x-y graph for t < e has a negative slope. This makes for a local maximum point.

If you were to write dy/dx as a function of x, then dy/dx would be pos. to the left of the critical point and would be neg to the right of it. I didn't do this calculation, but am pretty sure that would be the case.

 

[songoku]

Thank you very much Mark44

 

[Delta2]

Actually it isn't so hard to do it via the cartesian conversion , $$t=\sqrt{x^2-4}$$ $$y=\frac{\ln\sqrt{x^2-4}}{\sqrt{x^2-4}}$$ (and $x<-2$ since by the original parametric equations we can see that x is negative) and wolfram says that that function of $y(x)$ has a local maximum at $x=-\sqrt{4+e^2}$.