The question that I'm stuck on is this:
"Let f(x,y)= x^(2)*ln(y). Find the average rage of change of f as you go from (3,1) to (1,2) and find the instantaneous rate of change of f as you leave the point (3,1) toward (2,1)."
Any help would be appreciated. Thanks!
To find the average rate of change of a function f(x,y) as you go from point (x1, y1) to (x2, y2), you need to calculate the difference in the function values at the two points and divide it by the difference in the x and y values.
In this case, you are given the function f(x,y) = x^2 * ln(y) and two points, (3,1) and (1,2).
To find the average rate of change, let's calculate the difference in f(x,y) between the two points:
f(3,1) = 3^2 * ln(1) = 9 * ln(1) = 0
f(1,2) = 1^2 * ln(2) = ln(2)
Now calculate the differences in the x and y values:
Δx = 1 - 3 = -2
Δy = 2 - 1 = 1
Finally, find the average rate of change using the formula:
Average Rate of Change = Δf / (Δx + Δy)
Average Rate of Change = (ln(2) - 0) / (-2 + 1) = ln(2) / (-1) = -ln(2)
So, the average rate of change of f as you go from (3,1) to (1,2) is -ln(2).
Now, let's find the instantaneous rate of change of f as you leave the point (3,1) towards (2,1).
To find the instantaneous rate of change, we can take the partial derivative of f with respect to x and evaluate it at the point (3,1).
∂f/∂x = 2x * ln(y)
Now substitute x = 3 and y = 1 into the partial derivative:
∂f/∂x at (3,1) = 2(3) * ln(1) = 0
So, the instantaneous rate of change of f as you leave the point (3,1) toward (2,1) is 0.