So consider the following.
An insect is moving on the ellipse $2x^{2} + y^{2} = 3$ on the $xy$ – plane in the clockwise
direction at a constant speed of 3 centimeter per second.
The temperature function $T(x, y)$
(experienced by the insect) is given by
$T(x, y) = 3x^{2} − 2yx$,
where T is measured in degree Celsius and x, y are measured in centimeters. What is the rate
of change of the temperature (in degree Celsius per second) when the insect is at the point
(1, 1)?
So I more or less know how to do this question.
But the main obstacle I encounter is ensuring the tangent vector to the ellipse is actually pointing clock-wise. So how do I ensure the direction is clockwise ?
Best Answer
Parametrized the ellipse as $\displaystyle \mathbf{r}=\left( \frac{\sqrt{3}}{2}\sin \theta,\sqrt{3} \cos \theta \right)$ which is clockwise as $\theta$ increases.
$ds^{2}=dx^{2}+dy^{2}=\left( \frac{3}{2}\cos^{2} \theta+3\sin^{2} \theta \right) \, d\theta^{2}$
$\displaystyle \frac{d}{dt} T(x,y)= \frac{dT}{d\theta} \times \frac{d\theta}{ds} \times \frac{ds}{dt}= \frac{d}{d\theta} \left( \frac{9}{2}\cos^{2} \theta-3\sin \theta \cos \theta \right) \times \frac{1}{\sqrt{\frac{3}{2}\cos^{2} \theta+3\sin^{2} \theta}} \times v$
Then substitute $\displaystyle (\sin \theta,\cos \theta)= \left( \frac{2}{\sqrt{3}},\frac{1}{\sqrt{3}} \right)$ in your answer.
P.S.: If just simply find the tangent vector, let $f(x,y)=2x^{2}+3y^{2}$ then the outward normal is $$\mathbf{N}=\frac{\nabla f}{|\nabla f|}$$
Rotating clockwise by $90^{\circ}$, $$\mathbf{T}=(N_{y},-N_{x})$$