(a) To graph f(x, 0) against x, we substitute y = 0 into the function f(x, y). This gives us f(x, 0) = x^4 - 1. This is a simple quartic function and its graph will look like a "W" shape, with a minimum point at x = 0.

To graph f(0, y) against y, we substitute x = 0 into the function f(x, y). This gives us f(0, y) = y^2 - 1. This is a simple quadratic function and its graph will look like a "U" shape, with a minimum point at y = 0.

(b) The contour set f(x, y) = 3 is the set of points (x, y) that satisfy the equation x^4 + y^2 + 2x^2y - 1 = 3. This is a quartic equation in two variables and its graph will be a curve in the xy-plane.

(c) The directional derivative of f at the point (1, 1) in the direction of the vector (-1, 1) is given by the gradient of f at (1, 1) dotted with the unit vector in the direction of (-1, 1). The gradient of f is the vector of partial derivatives (4x^3 + 4xy, 2y + 2x^2), so at (1, 1) this is (8, 4). The unit vector in the direction of (-1, 1) is (-1/sqrt(2), 1/sqrt(2)), so the directional derivative is 8*(-1/sqrt(2)) + 4*(1/sqrt(2)) = -2*sqrt(2).

(d) The direction of steepest ascent at a point is given by the gradient at that point. So at (1, 1), the direction of steepest ascent is the vector (8, 4). The vector with length 1 in this direction is (8/sqrt(80), 4/sqrt(80)) = (4/sqrt(5), 2/sqrt(5)).

(e) The equation of the tangent plane at a point is given by f(x, y) = f(a, b) + f_x(a, b)*(x - a) + f_y(a, b)*(y - b), where (a, b) is the point and f_x and f_y are the partial derivatives of f. So at (1, 1), the equation of the tangent plane is f(x, y) = 3 + 8*(x - 1) + 4*(y - 1) = 8x + 4y - 9. The slope of steepest ascent is the magnitude of the gradient, which is sqrt(8^2 + 4^2) = sqrt(80), so the tangent plane is steepest in the direction of the gradient.

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(a) To graph f(x, 0) against x, we substitute y = 0 into the function f(x, y). This gives us f(x, 0) = x^4 - 1. This is a simple quartic function and its graph will look like a "W" shape, with a minimum point at x = 0.

To graph f(0, y) against y, we substitute x = 0 into the function f(x, y). This gives us f(0, y) = y^2 - 1. This is a simple quadratic function and its graph will look like a "U" shape, with a minimum point at y = 0.

(b) The contour set f(x, y) = 3 is the set of points (x, y) that satisfy the equation x^4 + y^2 + 2x^2y - 1 = 3. This is a quartic equation in two variables and its graph will be a curve in the xy-plane.

(c) The directional derivative of f at the point (1, 1) in the direction of the vector (-1, 1) is given by the gradient of f at (1, 1) dotted with the unit vector in the direction of (-1, 1). The gradient of f is the vector of partial derivatives (4x^3 + 4xy, 2y + 2x^2), so at (1, 1) this is (8, 4). The unit vector in the direction of (-1, 1) is (-1/sqrt(2), 1/sqrt(2)), so the directional derivative is 8*(-1/sqrt(2)) + 4*(1/sqrt(2)) = -2*sqrt(2).

(d) The direction of steepest ascent at a point is given by the gradient at that point. So at (1, 1), the direction of steepest ascent is the vector (8, 4). The vector with length 1 in this direction is (8/sqrt(80), 4/sqrt(80)) = (4/sqrt(5), 2/sqrt(5)).

(e) The equation of the tangent plane at a point is given by f(x, y) = f(a, b) + f_x(a, b)*(x - a) + f_y(a, b)*(y - b), where (a, b) is the point and f_x and f_y are the partial derivatives of f. So at (1, 1), the equation of the tangent plane is f(x, y) = 3 + 8*(x - 1) + 4*(y - 1) = 8x + 4y - 9. The slope of steepest ascent is the magnitude of the gradient, which is sqrt(8^2 + 4^2) = sqrt(80), so the tangent plane is steepest in the direction of the gradient.

Knowee AI · Accepted Answer

(a) To graph f(x, 0) against x, we substitute y = 0 into the function f(x, y). This gives us f(x, 0) = x^4 - 1. This is a simple quartic function and its graph will look like a "W" shape, with a minimum point at x = 0.

To graph f(0, y) against y, we substitute x = 0 into the function f(x, y). This gives us f(0, y) = y^2 - 1. This is a simple quadratic function and its graph will look like a "U" shape, with a minimum point at y = 0.

(b) The contour set f(x, y) = 3 is the set of points (x, y) that satisfy the equation x^4 + y^2 + 2x^2y - 1 = 3. This is a quartic equation in two variables and its graph will be a curve in the xy-plane.

(c) The directional derivative of f at the point (1, 1) in the direction of the vector (-1, 1) is given by the gradient of f at (1, 1) dotted with the unit vector in the direction of (-1, 1). The gradient of f is the vector of partial derivatives (4x^3 + 4xy, 2y + 2x^2), so at (1, 1) this is (8, 4). The unit vector in the direction of (-1, 1) is (-1/sqrt(2), 1/sqrt(2)), so the directional derivative is 8*(-1/sqrt(2)) + 4*(1/sqrt(2)) = -2*sqrt(2).

(d) The direction of steepest ascent at a point is given by the gradient at that point. So at (1, 1), the direction of steepest ascent is the vector (8, 4). The vector with length 1 in this direction is (8/sqrt(80), 4/sqrt(80)) = (4/sqrt(5), 2/sqrt(5)).

(e) The equation of the tangent plane at a point is given by f(x, y) = f(a, b) + f_x(a, b)*(x - a) + f_y(a, b)*(y - b), where (a, b) is the point and f_x and f_y are the partial derivatives of f. So at (1, 1), the equation of the tangent plane is f(x, y) = 3 + 8*(x - 1) + 4*(y - 1) = 8x + 4y - 9. The slope of steepest ascent is the magnitude of the gradient, which is sqrt(8^2 + 4^2) = sqrt(80), so the tangent plane is steepest in the direction of the gradient.

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