Find $\frac{d^2y}{dx^2}$ for Parametric Equations x,y

In summary, the given parametric equations find \frac{d^2y}{dx^2} as a function of t, for the given the parametric equations:\ x = 2-4*cos(t)y= 4+cos(t)^2\frac{d^2y}{dx^2} = -2*cos(t)*sin(t)dy/dt = -2*cos(t)*sin(t)second derv. 2*sin(t)^2-2*cos(t)^2dx/dt = 4*sin(t)second derv. 4*cos(t)\frac
  • #1
ILoveBaseball
30
0
find [tex]\frac{d^2y}{dx^2}[/tex] as a function of [tex]t[/tex], for the given the parametric equations:

x = 2-4*cos(t)
y= 4+cos(t)^2

[tex]\frac{d^2y}{dx^2}[/tex] = _______

dy/dt = [tex]-2*cos(t)*sin(t)[/tex]
second derv. [tex]2*sin(t)^2-2*cos(t)^2[/tex]

dx/dt = [tex]4*sin(t)[/tex]
second derv. [tex]4*cos(t)[/tex]

[tex]\frac{d^2y}{dx^2}[/tex] = [tex]\frac{2*sin(t)^2-2*cos(t)^2}{4*cos(t)}[/tex]

what did i do wrong?
 
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  • #2
[tex]\frac{d^2y}{dx^2} \neq \frac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}}[/tex]

Here is how I would do it.

[tex]x = 2 - 4\cos(t)[/tex]
[tex]4\cos(t) = 2-x[/tex]
[tex]t = \arccos\left(\frac{2-x}{4}\right)[/tex]

Now we have [tex]t[/tex] explicitly.

Substituite this back in for [tex]t[/tex] and we have

[tex]y = 4 + \cos\left(\arccos\left(\frac{2-x}{4}\right)\right)^2[/tex]
[tex]\sqrt{y} = 2 + \cos\left(\arccos\left(\frac{2-x}{4}\right)\right)[/tex]
[tex]\sqrt{y} = 2 + \left(\frac{2-x}{4}\right)[/tex]
[tex]\sqrt{y} = \frac{3}{2} - \frac{x}{4}[/tex]
[tex]y = \frac{9}{4} - \frac{x^2}{16}[/tex]

Now differentiate.

[tex]\frac{dy}{dx} = -\frac{x}{8}[/tex]
[tex]\frac{d^2y}{dx^2} = -\frac{1}{8}[/tex]
 
  • #3
Also in general, for y(t) and x(t),

[tex] \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} [/tex]

[tex] \frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2} [/tex]

Simplifying the RHS of the second equation you get:

[tex] \frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2} = \frac{d^2y}{dt^2} \frac{dt^2}{dx^2} [/tex] and by cancelling you get the desired result [tex] \frac{d^2y}{dx^2} [/tex]
 
  • #4
awesome thanks
 
  • #5
whozum said:
Also in general, for y(t) and x(t),

[tex] \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \ \ \ \color{blue} \longleftarrow \mathbf{ (EQ \ 1) } [/tex]

[tex] \color{red} \frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2} \ \ \ \mathbf{ \longleftarrow ( \underline{NOT\ \ CORRECT})} [/tex]

Simplifying the RHS of the second equation you get:

[tex] \frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2} = \frac{d^2y}{dt^2} \frac{dt^2}{dx^2} [/tex] and by cancelling you get the desired result [tex] \frac{d^2y}{dx^2} [/tex]
whozum:
Highlighted equation above for (d2y/dx2) is NOT correct. For parametric equations {y=y(t) & x=x(t)}, the second derivative is given by:

[tex] :(2): \ \ \ \ \ \color{blue} \frac{d^2y}{dx^2} \ = \ \left( \frac{d}{dt} \left( \frac{dy}{dx} \right) \right) / \left( \frac{dx}{dt} \right) [/tex]

where (dy/dx) is function of "t" obtained from the correct Equation #1 above.


~~
 
Last edited:

1. What is the formula for finding the second derivative of a parametric equation?

The formula for finding the second derivative of a parametric equation is d2y/dx2 = (d/dt)(dy/dt) / (dx/dt), where x and y are functions of t and t is the parameter.

2. Why do we need to find the second derivative of a parametric equation?

The second derivative of a parametric equation tells us the rate of change of the slope of the curve at any given point. This is important because it helps us understand the curvature of the curve and can provide information about the concavity and inflection points.

3. Can we use the chain rule to find the second derivative of a parametric equation?

Yes, we can use the chain rule to find the second derivative of a parametric equation. This is because the second derivative is the derivative of the first derivative, which is a function of t. Therefore, we can apply the chain rule to find the derivative of the first derivative with respect to t and then divide by the derivative of x with respect to t.

4. Is it possible for a parametric equation to have a second derivative that is undefined?

Yes, it is possible for a parametric equation to have a second derivative that is undefined. This can occur when the derivative of x with respect to t is equal to 0, which would result in a division by 0 when finding the second derivative. This can also happen when the curve has a sharp corner or cusp, where the slope changes abruptly and the second derivative does not exist at that point.

5. How does finding the second derivative of a parametric equation differ from finding the second derivative of a Cartesian equation?

The process of finding the second derivative of a parametric equation is similar to finding the second derivative of a Cartesian equation, as both involve taking the derivative of the first derivative. However, in a parametric equation, we must also divide by the derivative of x with respect to t, which is not necessary for Cartesian equations since x is already the independent variable. Additionally, parametric equations often require the use of the chain rule, while Cartesian equations may involve the use of implicit differentiation or the product rule.

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