Derivation of higher deratives of parametric equations

  • Thread starter thomas49th
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  • #1
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Homework Statement



so dy/dx = y dot / x dot

and d^2y/dx^2 = d/dx(y dot/x dot)

can someone please show me the steps to get

d^2y/dx^2 = (x dot y dot dot - y dot x dot dot)/ (x dot)^ 2

I've been trying to get to it for the last half an hour and failing

Thanks
 

Answers and Replies

  • #2
lanedance
Homework Helper
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so you have
[tex]\frac{dy}{dx} = \frac{y'}}{x'}} = y'(x')^{-1} [/tex]

now try differntiating using the product & chain rule
(i used dashes instead of dots, but same thing)
 
  • #3
655
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sorry but I don't see where you got x' = y'(x')^-1 from

Thanks
Thomas
 
  • #4
Cyosis
Homework Helper
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The product rule and chain rule need to be used here. Use the product rule first then use the chainrule. Example [itex]\frac{d}{dx} \frac{1}{y}=\frac{d}{dy}\left(\frac{1}{y}\right)\frac{dy}{dx}[/itex].

The first step is using the product rule on:

[tex]\frac{d}{dx} \frac{\dot{y}}{\dot{x}}[/tex].

Can you post the results?
 
Last edited:
  • #5
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sorry but I don't see where you got x' = y'(x')^-1 from

Thanks
Thomas

That isn't what he meant: he was writing out the equation and it started on a new line.

What he meant was:

[tex]\frac{dy}{dx} = \frac{y'}{x'} = y' \cdot (x')^{-1}[/tex]

and if you ask me you needn't do that step and head straight on to the quotient rule: [tex]\frac{vdu - udv}{v^{2}}[/tex] or [tex]\frac{vu'-uv'}{v^{2}}[/tex] where [tex]y' = u \ ; \ x' = v[/tex]

Just to expand on this:

[tex]\frac{du}{dx}=y''[/tex] or [tex]u'=y''[/tex]

[tex]\frac{dv}{dx}=x''[/tex] or [tex]v'=x''[/tex]

both give you the same result(duh) and there is not extra calculation needed.
 
Last edited:
  • #6
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I hope this has solved your problems.=]
 

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