Solving Parametric Equation: Find dy^2/dx^2 in Terms of t

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Discussion Overview

The discussion revolves around finding the second derivative \(\frac{d^2y}{dx^2}\) in terms of the parameter \(t\) for the parametric equations \(x = 2\cos t - \cos 2t\) and \(y = 2\sin t + \sin 2t\). Participants explore various methods of differentiation and express their results, seeking clarification and validation of their approaches.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents their result for \(\frac{d^2y}{dx^2}\) as \(\frac{1 + \cos t}{2\sin^3 t(1 - 2\cos t)}\) and questions its validity against a provided answer of \(\frac{-1}{\sin^3 t(2\cos t - 1)}\).
  • Another participant outlines the relationship between derivatives, stating \(\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\) and provides a formula for \(\frac{d^2y}{dx^2}\) involving second derivatives with respect to \(t\).
  • One participant mentions a method from their textbook, expressing a preference to stick with familiar techniques despite others suggesting alternative approaches.
  • There is a correction regarding the expression for \(\frac{dx}{dy}\), with participants clarifying the correct relationship between the derivatives.
  • Several participants share their calculations for \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), leading to differing expressions for \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\).

Areas of Agreement / Disagreement

Participants express differing results for \(\frac{d^2y}{dx^2}\) and debate the validity of their methods and calculations. No consensus is reached on the correct answer or the most effective approach to the problem.

Contextual Notes

Participants rely on various methods of differentiation and express uncertainty about their results. There are indications of missing assumptions or potential errors in calculations, but these remain unresolved.

naav
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Hi...i was just wondering if anyone gets the same answer to what i get for the following question...thanks...

find [tex]\frac{dy^2}{dx^2}[/tex] in terms of t for...

x = 2cost - cos2t, y = 2sint + sin2t...

i got my answer to be [tex]\frac{1 + cost}{2sin^3t(1 -2cost)}[/tex]

the answer is given as [tex]\frac{-1}{sin^3t(2cost -1)}[/tex]

have i gone wrong somewhere...do i need to simplify further...the answer i got and the one that is given do have some similarities so I'm just wondering...?...
 
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You have:
[tex]\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\to\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
Also:
[tex]\frac{d^{2}y}{dt^{2}}=\frac{d^{2}y}{dx^{2}}(\frac{dx}{dt})^{2}+\frac{dy}{dx}\frac{d^{2}x}{dt^{2}}[/tex]
Rearranging, we get:
[tex]\frac{d^{2}y}{dx^{2}}=\frac{\frac{d^{2}y}{dt^{2}}\frac{dx}{dt}-\frac{dy}{dt}\frac{d^{2}x}{dt^{2}}}{(\frac{dx}{dt})^{3}}[/tex]
 
Last edited:
Hi...thank you...

i found:

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

= (d/dt)(dy/dx) * (dt/dx)

did you get the same answer as me...?...
 
Do not use that form; instead let:
y(t)=y(x(t))
Follow the derivation in post 3 to find the correct expression.
(The notation used here is rather sloppy, but it shouldn't be too difficult to follow)
 
Hi...thank you...

the method i mentioned is the way we've been shown in the textbook and in the notes...and that's the way I've tackled other questions...it's better for me to stick to the method shown in the classes...

i got dx/dt = -2sint + 2sin2t and dy/dt = 2cost + 2cos2t...

then i got dy/dx = (cost + 1)/sint

then (d^2t)/(dx^2) = (d/dt)(dy/dx) * dt/dx...

i got (d/dt)(dy/dx) = (-1 - cost)/(sin^2t)

and dt/dx = 1/(-2sint + 2cost)

then (d^2t)/(dx^2) = (1 + cost)/[(2sin^3(t)(1 - cost)]...

but the answer is given as something else...?...
 
naav said:
the method i mentioned is the way we've been shown in the textbook and in the notes...and that's the way I've tackled other questions...it's better for me to stick to the method shown in the classes...
Personally, I would stick with the more efficient/clever method regardless of where I've learned it from, but anywho...

i got dx/dt = -2sint + 2sin2t and dy/dt = 2cost + 2cos2t...
Correct.
then i got dy/dx = (cost + 1)/sint
Where are you getting this from? You might one to check this again.
 
Hi...thank you...

i got dx/dy = (dy/dt)/(dx/dt)...

according to the answer I've got that bit right...
 
naav said:
Hi...thank you...

i got dx/dy = (dy/dt)/(dx/dt)...

according to the answer I've got that bit right...

No, dx/dy= (dx/dt)/(dy/dt)
 
Hi...sorry, i was meaning the first derivative - dy/dx...

dy/dx = (dy/dt)/(dx/dt)
 

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