Past exam of the mathematical methods II course of the University of Oxford / 2013 / 4

The gradient is $\nabla T=2(x,y)/(x^2+y^2)$ and at $(1,2)$ it is $(2/5,4/5)$. The directional derivative along a unit vector is maximal in the gradient direction. ThereforeThe direction is radially outward from the origin, perpendicular to the circular level curve through the point.

Let $s\ge 0$ be distance along the unit direction from part (a). The point is $\mathbf{r}(s)=(1,2)+s(1,2)/\sqrt{5}$, whose distance from the origin is $\sqrt{5}+s$. For finite increments of a logarithmic radial function, the exact change isThus an increase of $0.2$ requiresIf the intended calculation is the local directional derivative estimate, its first-order value is $s\simeq \Delta T/|\nabla T|=\boxed{\sqrt{5}/10}$. The latter approximates the exact finite distance rather than equaling it. Numerically they are about $0.235$ and $0.224$, respectively.

Normalize the specified direction to $\hat{\mathbf{u}}=(1,1)/\sqrt{2}$. At distance $s\ge 0$ the squared radius is $(1+s/\sqrt{2}{)}^2+(2+s/\sqrt{2}{)}^2=5+3\sqrt{2}s+s^2$. The exact change $\Delta T=0.12$ therefore requires $5+3\sqrt{2}s+s^2=5e^{0.12}$. The positive root isFor comparison, the initial directional derivative is $(2/5,4/5)\cdot (1,1)/\sqrt{2}=3\sqrt{2}/5$. Its first-order distance estimate isThese give approximately $0.145$ and $0.141$. The exact expression accounts for the changing gradient along a finite straight displacement.

A stationary direction at the point has zero directional derivative, hence is perpendicular to $(1,2)$. The two opposite unit directions areThey are tangents to the circular level curve. “Stationary” here means zero first-order change. A finite straight move in a tangent direction gives squared radius $5+s^2$ and therefore a positive second-order change $\ln (1+s^2/5)$; following the curved level set instead keeps $T$ exactly constant.