Multivariable Calculus, rate of change.

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 $(1,1)$?

Hint: Let $f(x;y)=2x^2+y^2$. The gradient vector of $(1,1) = \langle4, 2 \rangle$ is normal to the ellipse $f(x,y)=3$ at the point $(1,1)$. Using this information, how can we easily find a vector which is tangential to the ellipse $f(x,y)=3$ and is pointing in the clockwise direction?

My attempt to this question is firstly, since the gradient vector is a normal vector to the ellipse, I can use the property of it being a normal vector, such that the tangential vector, represented by $t$, $n \cdot t = 0$. However, is it possible to just use any values such that $tx$ and $ty$ can fulfill this property? After which, how can I proceed from here!

Let $\vec{v}=\langle4,2\rangle$. Then a vector orthogonal to $\vec{v}$ can be found by interchanging the components and changing one of the signs, so we can take $\vec{t}=\langle2,-4\rangle$ to get a tangent vector corresponding to the clockwise direction; and normalizing this vector gives the unit tangent vector $\vec{u}=\langle\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}\rangle$.

Now find the directional derivative of T in the direction of $\vec{u}$ using

$\;\;\;D_{\vec{u}}T(1,1)=\vec{\nabla T}(1,1)\cdot\vec{u}$

The ellipse $\gamma:\> 2x^2+y^2=3$ is a level curve of your function $f(x,y):=2x^2+y^2$. Therefore $\nabla f(1,1)=(4,2)$ is orthogonal to the tangent of $\gamma$ at $(1,1)$. It follows that one of $\pm(-2,4)$ is the tangent vector you are after. Since the movement of the insect is clockwise a figure immediately tells us that we should choose $v_*:=(2,-4)$. The actual velocity vector $v$ of the insect is a positive multiple of $v_*$, but has absolute value $3$. It follows that $$v=3{(2,-4)\over\sqrt{20}}=\left({3\over\sqrt{5}}, \>-{6\over\sqrt{5}}\right)\ .$$ The temperature $u(t)$ felt by the insect at time $t$ is given by $$u(t)=T\bigl(x(t),y(t)\bigr)\ .$$ According to the chain rule we then have $$\dot u(t)=\nabla T\bigl(x(t),y(t)\bigr)\cdot\bigl(\dot x(t),\dot y(t)\bigr)\ .$$ Evaluating this at the moment $t_0$ where the insect passes the point $(1,1)$ we obtain $$\dot u(t_0)=\nabla T(1,1)\cdot v=(4,-2)\cdot\left({3\over\sqrt{5}}, \>-{6\over\sqrt{5}}\right)={24\over\sqrt{5}}\ .$$

• why do you have to multiply by 3 to the vector $v_*$? I pretty did everything you did but without that step. I'm unclear of why you need to do that – Danxe Oct 8 '15 at 12:43
• @Danxe: Because we are told that the speed of the fly is $3$ cm per second. – Christian Blatter Oct 8 '15 at 12:54
• Is there a particular reason why we need to factor in the actual velocity? Sorry if i'm asking so much, i can't seem to understand. – Danxe Oct 8 '15 at 13:10
• @Danxe: If the fly doubles its speed from $3$ cm/sec to $6$ cm/sec the felt temperature will increase twice as fast. – Christian Blatter Oct 8 '15 at 13:13

you can turn this problem of two variables into one variable problem if you parametrise the ellipse as $$x = \sqrt{\frac32} \cos \theta, y = \sqrt3\sin \theta.$$

now we can find the angular velocity at $x = 1, y= 1, \cos\theta = \sqrt{\frac23}, \sin\theta=\sqrt\frac13$ if we set $$ds = \sqrt{dx^2 + dy^2} = \sqrt{(\frac32\sin^2\theta + 3\cos^2 \theta)}\,d\theta=\sqrt{\frac12+2}d\theta=\sqrt\frac52d\theta.$$

therefore at $$x = y = 1,\frac{d\theta}{dt} = \sqrt\frac52, dx=-\sqrt\frac32\sin\theta d\theta=-\sqrt\frac12 d\theta, dy = \sqrt3\cos \theta d\theta=\sqrt 2 d \theta$$

now $$dT = d(3x^2 - 2xy)=6xdx-2ydx-2xdy =6dx-2dx-2dy=-4\sqrt2d\theta$$

therefore, $$\frac{dT}{dt} = -4\sqrt 5$$