Final answer:
To find the slope of the tangent line to f(x,y) = x³y-y² when x = 1 and y = 2, in the direction of x = 4 and y = 3, we need to find the partial derivatives of the function with respect to x and y. Substituting the given values into the partial derivatives, we find f_x(1,2) = 6 and f_y(1,2) = -3. By taking the dot product of the direction vector (3, 1) with the gradient vector (6, -3), we find the slope of the tangent line to be 15.
Step-by-step explanation:
To find the slope of the tangent line to f(x,y) = x³y-y² when x = 1 and y = 2, in the direction of x = 4 and y = 3, we first need to find the partial derivatives of the function with respect to x and y.
The partial derivative of f(x,y) with respect to x is given by fx(x,y) = 3x²y. The partial derivative of f(x,y) with respect to y is given by fy(x,y) = x³-2y.
Next, we substitute the given values of x and y into the partial derivatives:
fx(1,2) = 3(1)²(2) = 6
fy(1,2) = (1)³-2(2) = -3
The slope of the tangent line in the direction of x = 4 and y = 3 can be found by taking the dot product of the direction vector ⋂ (4-1, 3-2) = (3, 1) with the gradient vector (∇f)(1,2) = (6, -3):
Slope = (3, 1) • (6, -3) = 3(6) + 1(-3) = 18 - 3 = 15
Therefore, the slope of the tangent line to f(x,y) when x = 1 and y = 2, in the direction of x = 4 and y = 3 is 15.