Vector Fields — Question 6

Question 6 – Solution

Strategy. The gradient is normal to level curves; multiplying it by a negative constant reverses direction but preserves normality.

Step 1: Construct the field $$\nabla T=\langle-2x,-8y\rangle, \qquad \mathbf H=-\frac 12\nabla T =\boxed{\langle x,4y\rangle}.$$

Step 2: Evaluate at $P$ $$\mathbf H(2,1)=\langle 2,4\rangle,\qquad |\mathbf H(2,1)|=\boxed{2\sqrt 5},$$ $$\widehat{\mathbf H(2,1)} =\boxed{\left\langle\frac 1{\sqrt 5},\frac 2{\sqrt 5}\right\rangle}.$$

Step 3: Isotherm and normality Since $T(2,1)=100-4-4=92$, the isotherm is $$x^2+4y^2=8.$$ Its normal direction is $\nabla(x^2+4y^2)=\langle 2x,8y\rangle$. At $P$ this is $\langle 4,8\rangle=2\mathbf H(P)$, so the flux is normal to the ellipse and points toward lower temperature.

Verification Moving a short distance along $\mathbf H(P)$ increases $x^2+4y^2$, so it decreases $T=100-(x^2+4y^2)$, as heat flow should.