Finding d^2y/dx^2 for a parametrized curve

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Therefore,\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{1}{1-2t}\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{1}{1-2t}\frac{d}{dt}\left(\frac{dy}{dt}\frac{dt}{dx}\right)=\frac{1}{1-2t}\left(\frac{d^2y}{dt^2}\frac{dt}{dx}-\frac{dy}{dx}\frac{d}{dx}\right)Now
  • #1
IntegrateMe
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Find d2y/dx2 as a function of t if x = t - t2 and y = t - t3

Ok, so this basically means we're going to take the second derivative of y with respect to t over the derivative of x with respect to t.

So the second derivative of y would be.

[2 - 6t + 6t2]/(1 - 2t)2

Now, we just put that over the first derivative of x which is 1 - 2t so we get:

[[2 - 6t + 6t2]/(1 - 2t)2]/(1 - 2t)

=

(2 - 6t + 6t2)/(1 - 2t)3

Correct or incorrect?
 
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  • #2
IntegrateMe said:
Find d2y/dx2 as a function of t if x = t - t2 and y = t - t3

Ok, so this basically means we're going to take the second derivative of y with respect to t over the derivative of x with respect to t.

Why do you think this?

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

Use the chain rule

[tex]\frac{d}{dx}=\frac{dt}{dx}\frac{d}{dt}=\frac{1}{x'(t)}\frac{d}{dt}[/tex]
 

1. What is the definition of d^2y/dx^2 for a parametrized curve?

The second derivative of a parametrized curve is a measure of the curvature of the curve at a specific point. It represents the rate of change of the slope of the curve with respect to the x-coordinate at that point.

2. How do you find d^2y/dx^2 for a parametrized curve?

To find the second derivative of a parametrized curve, you first need to find the first derivative dy/dx. Then, you can apply the chain rule to find the second derivative, which is expressed as (d^2y/dt^2)/(dx/dt)^2.

3. What is the significance of finding d^2y/dx^2 for a parametrized curve?

The second derivative provides important information about the shape and behavior of a parametrized curve. It can help determine the presence of critical points, points of inflection, and concavity of the curve.

4. Can d^2y/dx^2 be negative or zero for a parametrized curve?

Yes, the second derivative of a parametrized curve can be negative or zero at certain points. A negative second derivative indicates a concave down curve, while a zero second derivative indicates a point of inflection.

5. Are there any applications of finding d^2y/dx^2 for a parametrized curve?

Yes, the second derivative of a parametrized curve has various applications in fields such as physics, engineering, and economics. It can help analyze the motion of objects, optimize functions, and model economic trends.

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