MHB Why Does dy/dt Equal Zero at the Endpoint of the Minor Axis?

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The discussion centers on the derivative of the equation (x-1)² + 2y² = 2, specifically exploring why dy/dt equals zero at the endpoint of the minor axis. The calculation shows that dy/dt is indeed zero at the point (1,1), indicating a transition in the behavior of y from increasing to decreasing with respect to time. Participants clarify that this result does not imply that y is not changing overall, but rather that its rate of change is zero at that specific point. The conversation highlights the importance of understanding implicit differentiation in relation to motion along an elliptical path. Ultimately, the conclusion reinforces that dy/dt can be zero at critical points in a dynamic system.
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(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 …

$\theta = \arctan\left(\dfrac{y}{x}\right)$
 
skeeter said:
(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 …

$\theta = \arctan\left(\dfrac{y}{x}\right)$

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...
 
zimsam 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...

is that so … ?

AF0BC4B2-A1E4-4189-AA13-A9981948D0CA.png
 
skeeter said:
is that so … ?

View attachment 11130

I already know that dy/dt must be changing as well...
How did I make a mistake in my implicit differentiation?
 
zimsam said:
I already know that dy/dt must be changing as well...
How did I make a mistake in my implicit differentiation?

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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