Checking nature of turning point of parametric equation

In summary, the turning point is a local maximum when dy/dx is positive to the left and negative to the right.
  • #1
songoku
2,319
331
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
 
Physics news on Phys.org
  • #2
songoku said:
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.
 
  • Like
Likes songoku
  • #3
Mark44 said:
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
 
  • #4
songoku said:
dy/dx in paramateric form or in cartesian form?
Thanks
I calculated dy/dx as a function of the parameter, t.
 
  • Like
Likes songoku
  • #5
Mark44 said:
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
 
  • #6
songoku said:
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.
 
  • Like
Likes songoku
  • #7
Thank you very much Mark44
 
  • #8
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}##.
 
  • Like
Likes songoku

Related to Checking nature of turning point of parametric equation

1. What is a turning point in a parametric equation?

A turning point in a parametric equation is a point where the direction of the curve changes, either from increasing to decreasing or vice versa.

2. How do you determine the nature of a turning point in a parametric equation?

To determine the nature of a turning point in a parametric equation, you can find the second derivative of the equation and evaluate it at the turning point. If the second derivative is positive, the turning point is a local minimum. If it is negative, the turning point is a local maximum. If the second derivative is zero, further analysis is needed to determine the nature of the turning point.

3. Can a turning point in a parametric equation be a point of inflection?

Yes, a turning point in a parametric equation can also be a point of inflection. A point of inflection is a point where the curve changes from being concave upwards to concave downwards, or vice versa.

4. What is the significance of determining the nature of a turning point in a parametric equation?

Determining the nature of a turning point in a parametric equation can help in understanding the behavior of the curve and making predictions about its future behavior. It can also be useful in optimization problems where the goal is to find the maximum or minimum value of a function.

5. Are there any other methods for determining the nature of a turning point in a parametric equation?

Yes, there are other methods such as using the first derivative test or graphing the parametric equation to visually analyze the turning point. However, finding the second derivative and evaluating it at the turning point is the most commonly used method.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
393
  • Calculus and Beyond Homework Help
Replies
8
Views
638
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
793
  • Calculus and Beyond Homework Help
Replies
11
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
968
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Back
Top