Question

Find all points at which the direction of fastest change of the function \(f(x,y) - {x^2} + {y^2} - 2x - 4y\) is\({\bf{i}} + {\bf{j}}\).

Short answer

The direction of fastest change is \(i + j\) for all points on the line\(y = x + 1\).

Step-by-step solution

Calculate the direction of fastest change

The function is\(f(x,y) = {x^2} + {y^2} - 2x - 4y\).

Find the normal vector \(\nabla f(x,y)\) of the function\(f(x,y) = {x^2} + {y^2} - 2x - 4y\).

\(\begin{aligned}{l}\nabla f(x,y) = \frac{d}{{dx}}i + \frac{d}{{dy}}j\\ = \frac{\partial }{\partial }\left( {{x^2} + {y^2} - 2x - 4y} \right)i + \frac{\partial }{{\partial y}}\left( {{x^2} + {y^2} - 2x - 4y} \right)j\\ = (2x - 2)i + (2y - 4)j\end{aligned}\)

The direction of fastest change is \((2x - 2)i + (2y - 4)j\) and it should be parallel to \(i + j\)

Therefore,

\(\begin{aligned}{l}Vf(x,y) = k(i + j)\\(2x - 2)i + (2y - 4)j = k(i + j)\\k = 2x - 2,k = 2y - 4\end{aligned}\)

Then, equate the \(k\) equation,

\(\begin{aligned}{l}2x - 2 = 2y - 4\\2x - 2y - 2 + 4\\02x + 2 = 2y\\y = x + 1\end{aligned}\)

Therefore, it can be concluded that the direction of fastest change is \(\vec i + j\) for all points on the line\(y = x + 1\).