- #1
zimsam
- 3
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Why Does dy/dt Equal Zero at the Endpoint of the Minor Axis?
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Discussion Overview
The discussion revolves around the mathematical reasoning behind why the derivative $\frac{dy}{dt}$ equals zero at the endpoint of the minor axis of a given equation, specifically $(x-1)^2 + 2y^2 = 2$. Participants explore the implications of this result in the context of implicit differentiation and the behavior of the variables involved.
Discussion Character
- Technical explanation
- Mathematical reasoning
- Debate/contested
Main Points Raised
- Some participants present the equation $(x-1)^2 + 2y^2 = 2$ and suggest using implicit differentiation to find $\frac{dy}{dt}$ given $\frac{dx}{dt}$.
- One participant calculates $\frac{dy}{dt}$ and finds it to be zero, questioning the validity of this result by asserting that the rate of change of $y$ cannot be zero.
- Another participant expresses confusion about their implicit differentiation process, seeking clarification on their potential mistake.
- A later reply indicates that $\frac{dy}{dt}$ equals zero at the point (1,1), which is identified as the endpoint of the minor axis, and notes that $y$ transitions from increasing to decreasing at that position.
Areas of Agreement / Disagreement
Participants express differing views on the implications of $\frac{dy}{dt}$ being zero. While some acknowledge that this is a valid outcome at the endpoint, others contest the interpretation of what this means for the behavior of $y$ over time.
Contextual Notes
There are unresolved questions regarding the assumptions made during implicit differentiation and the specific conditions under which $\frac{dy}{dt}$ is evaluated. The discussion reflects uncertainty about the implications of the mathematical results presented.
- #2
skeeter
- 1,103
- 1
(a) $\dfrac{d}{dt}\bigg[(x-1)^2+ 2y^2=2 \bigg]$
you are given $\dfrac{dx}{dt}$ and the position of the planet.
use the derivative to calculate $\dfrac{dy}{dt}$
(b) hint …
$\theta = \arctan\left(\dfrac{y}{x}\right)$
you are given $\dfrac{dx}{dt}$ and the position of the planet.
use the derivative to calculate $\dfrac{dy}{dt}$
(b) hint …
$\theta = \arctan\left(\dfrac{y}{x}\right)$
- #3
zimsam
- 3
- 0
skeeter said:
When I find dy/dt, it comes out to =0 for me. dy/dx=(-dx/dt(x-1))/2y. Surely the rate of change of y is not zero...
- #4
skeeter
- 1,103
- 1
zimsam said:
is that so … ?
- #5
zimsam
- 3
- 0
skeeter said:
I already know that dy/dt must be changing as well...
How did I make a mistake in my implicit differentiation?
- #6
skeeter
- 1,103
- 1
zimsam said:
you didn’t make a mistake …
dy/dt = 0 at (1,1) which is at the endpoint of the minor axis
y is changing from increasing to decreasing w/respect to time at that position
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