Concavity of Parametric Equations

In summary, the problem asks to find dy/dx and d^y/dx^2 for the given parametric equations x = e^t, y = te^(-t). The question then asks for the values of t where the curve is concave upward. To find this, the formula for d^2y/dx^2 is used and set to zero. After solving, the correct answer is t>3/2. In a similar problem with different parametric equations x = cos 2t, y = cos t, the same process is used but without the incorrect exponential term.
  • #1
sheldonrocks97
Gold Member
66
2

Homework Statement



Find dy/dx and d^y/dx^2

x=e^t; y=te^(-t)

For which values of t are concave upward? (write your answer in interval notation).

Homework Equations


The Attempt at a Solution



I used the formula to find d^2y/dx^2.

d^2y/dx^2= e^(-3t)*(2t-3)

Set it to zero:

e^(3t)*(2t-3)>0

I solved it and got t>3/2, but the computer told me it was wrong. What am I doing wrong here?
 
Last edited:
Physics news on Phys.org
  • #2
sheldonrocks97 said:

Homework Statement



Find dy/dx and d^y/dx^2

x = cos 2t, y = cos t, 0 < t < ∏

For which values of t are concave upward? (write your answer in interval notation).

Homework Equations





The Attempt at a Solution



I used the formula to find d^2y/dx^2.

d^2y/dx^2= e^(-3t)*(2t-3)

I solved it and got t>3/2, but the computer told me it was wrong. What am I doing wrong here?

Where did the exponential come from? You only have cosines in the problem. After you fix that, please show your work so we can follow it.
 
  • #3
LCKurtz said:
Where did the exponential come from?
The OP edited his post, so the parametric equations now have exponential form
LCKurtz said:
After you fix that, please show your work so we can follow it.
Yes. Show your work for dx/dt and dy/dt and so on.
 

1. What is concavity of parametric equations?

The concavity of parametric equations refers to the shape or curvature of a graph created by plotting the x and y coordinates of a set of parametric equations. It can be described as either concave up (curving upward) or concave down (curving downward).

2. How is concavity determined for parametric equations?

The concavity of parametric equations is determined by calculating the second derivative of the equations with respect to the independent variable (usually t). If the second derivative is positive, the graph is concave up, and if it is negative, the graph is concave down.

3. What is the significance of concavity in parametric equations?

The concavity of parametric equations can provide important information about the behavior and characteristics of the graph. For example, concave up graphs tend to have a minimum point, while concave down graphs tend to have a maximum point. Concavity can also affect the rates of change and the direction of the graph.

4. Can parametric equations have changing concavity?

Yes, parametric equations can have changing concavity. This occurs when the second derivative changes sign, resulting in a change in the direction of the curvature. This can be observed on the graph as a point where the graph changes from being concave up to concave down, or vice versa.

5. How can concavity of parametric equations be used in real-world applications?

The concept of concavity in parametric equations can be applied in various fields such as physics, engineering, and economics. For example, it can be used to analyze the trajectory of a projectile, the shape of a bridge, or the cost and revenue functions of a business. Understanding the concavity can help predict the behavior and make important decisions based on the graph.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
392
  • Calculus and Beyond Homework Help
Replies
2
Views
514
Replies
2
Views
919
  • Calculus and Beyond Homework Help
Replies
20
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
854
  • Calculus and Beyond Homework Help
Replies
5
Views
622
  • Calculus and Beyond Homework Help
Replies
5
Views
733
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
665
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
Back
Top