Parametric Equation Homework: Show Constant of ##\frac{d^2y}{dx^2}/(dy/dx)^4##

  • Thread starter squenshl
  • Start date
  • Tags
    Parametric
In summary, to show that ##\frac{\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^4}## is a constant, differentiate both parametric equations with respect to ##t##, apply the chain rule, and then substitute in the resulting expressions for the second derivative and first derivative. Keep in mind the chain rule when differentiating the second derivative.
  • #1
squenshl
479
4

Homework Statement


A curve is defined by the parametric equations ##x=t^3+1## and ##y=t^2+1##.
Show that ##\frac{\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^4}## is a constant.

Homework Equations

The Attempt at a Solution


So you differentiate both equations wrt ##t## then apply the chain rule to get ##\frac{2}{3t}##. Applying the chain rule after differenating twice to get ##\frac{d^2y}{dx^2}=\frac{1}{3t}##.
Substitute in both to get the result?
 
Physics news on Phys.org
  • #2
squenshl said:

Homework Statement


A curve is defined by the parametric equations ##x=t^3+1## and ##y=t^2+1##.
Show that ##\frac{\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^4}## is a constant.

Homework Equations

The Attempt at a Solution


So you differentiate both equations wrt ##t## then apply the chain rule to get ##\frac{2}{3t}##. Applying the chain rule after differenating twice to get ##\frac{d^2y}{dx^2}=\frac{1}{3t}##.
Substitute in both to get the result?
Yes. Keep in mind that ##\frac{d^2 y}{dx^2} = \frac d {dx} \left(\frac {dy}{dx}\right) \cdot \frac {dt}{dx}##, using the chain rule.
 
  • #3
Mark44 said:
Yes. Keep in mind that ##\frac{d^2 y}{dx^2} = \frac d {dx} \left(\frac {dy}{dx}\right) \cdot \frac {dt}{dx}##, using the chain rule.
Got it thanks a lot
 
Last edited:

1. What is a parametric equation?

A parametric equation is a mathematical representation of a curve or surface in terms of one or more parameters, which are independent variables that determine the position of points on the curve or surface.

2. How do you find the constant in the equation ##\frac{d^2y}{dx^2}/(dy/dx)^4##?

The constant in this equation is the coefficient of the highest power of the dependent variable, y. To find it, you can first rewrite the equation as ##\frac{d^2y}{dx^2} * (dy/dx)^{-4}##. Then, you can use the quotient rule to take the derivative of this expression with respect to x. The resulting constant will be the coefficient of the highest power of y.

3. What is the significance of the constant in the parametric equation?

The constant in a parametric equation represents a fixed value that is independent of the parameters. In other words, it is a constant that remains the same regardless of the values of the parameters. This can be useful in certain applications, such as modeling physical systems or analyzing data.

4. Can you provide an example of a parametric equation?

One example of a parametric equation is the equation of a circle in two dimensions, which can be expressed as x = r*cos(t) and y = r*sin(t), where r is the radius of the circle and t is the parameter that determines the position of points on the circle. This equation represents a circle with center (0,0) and radius r.

5. How is a parametric equation different from a Cartesian equation?

A parametric equation uses parameters to define the coordinates of points on a curve or surface, while a Cartesian equation uses variables to define the relationship between x and y coordinates. In other words, a parametric equation describes the motion of a point, while a Cartesian equation describes the shape of a curve or surface.

Similar threads

  • Calculus and Beyond Homework Help
Replies
20
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
462
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
758
  • Calculus and Beyond Homework Help
Replies
7
Views
706
  • Calculus and Beyond Homework Help
Replies
3
Views
273
  • Calculus and Beyond Homework Help
Replies
3
Views
819
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
Back
Top