Derivative of a parametric equation

  • Thread starter Karol
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  • #1
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Homework Statement


$$y=1+t^2,~~x=\frac{t}{1+t^2}$$
What is dy/dx

Homework Equations


Parametric equation's derivative:
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

The Attempt at a Solution


$$\frac{dx}{dt}=\frac{1-t^2}{(1+t^2)^2}$$
$$\frac{dy}{dx}=\frac{2t(1+t^2)^2}{1-t^2}$$
I can't translate it back to x
 

Answers and Replies

  • #2
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Why do you want to do it? Anyway, you could also start from ##t=x\cdot y## and express ##\frac{dy}{dx}## as a function of ##x## and ##y##.
 
  • #3
Dick
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Homework Helper
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Homework Statement


$$y=1+t^2,~~x=\frac{t}{1+t^2}$$
What is dy/dx

Homework Equations


Parametric equation's derivative:
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

The Attempt at a Solution


$$\frac{dx}{dt}=\frac{1-t^2}{(1+t^2)^2}$$
$$\frac{dy}{dx}=\frac{2t(1+t^2)^2}{1-t^2}$$
I can't translate it back to x

The derivative won't be a function only of ##x##. Use that ##t=xy## if you want to express it as a function of ##x## and ##y##.
 
  • #4
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The derivative won't be a function only of ##x##.
Why? indeed i cannot express t as a function of x, is that the reason?
 
  • #5
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Why? indeed i cannot express t as a function of x, is that the reason?
Right. To solve for t in the equation ##x = \frac t {1 + t^2}##, you would most likely use the quadratic formula, which will give two values of t. So t is not a function of x.
 
  • #6
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Thank you fresh_42, Dick and Mark44
 
  • #7
Ray Vickson
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Why? indeed i cannot express t as a function of x, is that the reason?

You have already been told (in another, similar, thread) that solving for ##t## in terms of ##x## will give you two different formulas, so will give you two different curves ##y = f_1(x)## and ##y = f_2(x)##. That means you will get two different derivative formulas for ##dy/dx##.
 

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