Question

Find the maximum rate of change of f(x,y)=ln(x2+y2) at the point (-2, 2) and the direction in which it occurs.

Answer 1

Answer:

Step-by-step explanation:

Maximum rate of change of a function f(x,y) is expressed as the gradient of such function at the given point. Gradient of a function is expressed as;

∇f(x,y) = $\dfrac{\delta f}{\delta x}i + \dfrac{\delta f}{\delta y}j$

Given f(x,y)=ln(x²+y²) at the point (-2, 2)

$\dfrac{\delta f}{\delta x} = 2x * \frac{1}{x^2+y^2} \\\dfrac{\delta f}{\delta x} = \frac{2x}{x^2+y^2} \\\\\\\dfrac{\delta f}{\delta y} = 2y * \frac{1}{x^2+y^2} \\\dfrac{\delta f}{\delta y} = \frac{2y}{x^2+y^2}$

$\nabla f(x, y) = \frac{2x}{x^2+y^2}i+\frac{2y}{x^2+y^2}j\\ \\\nabla f(x, y) \ at \ the \ point \ (-2,2) \ will \ give;\\\nabla f(x, y) = \frac{2(-2)}{(-2)^2+2^2}i+\frac{2(2)}{(-2)^2+2^2}j\\ \nabla f(x, y) = \frac{-4}{8}i+\frac{4}{8}j\\\nabla f(x, y) = \frac{-1}{2}i+\frac{1}{2}j$

Hence the maximum rate of range of the function at the point (-2,2) is $\frac{-1}{2}i+\frac{1}{2}j$

To get the direction, we willl use the formula;

$tan \theta = \frac{y}{x}\\ \theta = tan^{-1}\frac{y}{x}$

From the vector gotten, x = -1/2 and y = 1/2

$\theta = tan^{-1}\frac{(1/2)}{-1/2}\\ \theta = tan^{-1}-1\\ \theta = -45^0$

Since tan theta is negative in the second quadrant, the direction in which it occurs will be 180-45 = 135°