Question

The temperature at a point (x,y) on a rectangular metal plate is given by T(x,y) = 100 - 2x^2 - y^2. Find the path a heatseeking particle will take, starting at (3,4) as it moves in the direction in which the temperature increases the most rapidly.

Since we want the path to move in gradient of T then path = x'(t) + y'(t).

gradient of T = -4x i - 2y j

Solving for dx/dt = -4x , dx/x = -4dt

… now I'm stuck. I don't really know how to go from here.

Answer 1

\[T(x,y)=100-2x^2-y^2\]\[\nabla T(x,y)=-4x\vec{i}-2y\vec{j}\]\[\mbox{Path: }y=f(x)\]\[y'=\frac{-2y}{-4x}=\frac{y}{2x}\quad\Rightarrow\quad\frac{y'}{y}=\frac{1}{2x}\quad\Rightarrow\quad (\ln{y})'=\frac{1}{2x}\]\[\ln{y}=\int\frac{dx}{2x}\quad\Rightarrow\quad \ln{y}=\frac{1}{2}\ln{x}+C=\ln{\sqrt{x}}+C\]\[y=\exp{(\ln{\sqrt{x}}+C)=C_1\sqrt{x}}\]\[4=C_1\sqrt{3}\quad\Rightarrow\quad C_1=4/\sqrt{3}\]\[\fbox{$y=4\sqrt{\frac{x}{3}}$}\]

Answer 2

\[\mbox{parabola: }x=\frac{3}{16}y^2\quad,\quad y\geq 0\]

Answer 3

Interesting. How did you got from ln y=1/2 ln x+C to ln √x+C ?

Answer 4

\[\ln{x^\alpha}=\alpha\ln{x}\]

Answer 5

Thanks, also what is exp() ?

Answer 6

\[\exp{(x)}=e^x\]

Answer 7

Wouldn't that give\[\sqrt{x}e^C\] or are these different Cs?