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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 ?

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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 $90^{\circ}$, $$\mathbf{T}=(N_{y},-N_{x})$$

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