Directional Derivatives & Steepest Ascent

The gradient is the vector of partial derivatives (∂f/∂x, ∂f/∂y). To find ∂f/∂x, we treat y as a constant. The derivative of x is 1, and sin(y) is a constant coefficient, so ∂f/∂x = 1*sin(y) = sin(y). To find ∂f/∂y, we treat x as a constant. The derivative of sin(y) is cos(y), so ∂f/∂y = x cos(y). Therefore, ∇f = (sin(y), x cos(y)).

The magnitude of the gradient represents the steepness or the rate of change of the function. On a contour map, if level curves (representing equal steps in height/value) are packed closely together, it means the function value is changing rapidly over a short distance. This corresponds to a steep slope and, consequently, a gradient vector with a large magnitude.

The gradient ∇f is the vector of partial derivatives (∂f/∂x, ∂f/∂y). To find ∂f/∂x, we treat y as a constant and differentiate x³y² with respect to x. Using the power rule, the derivative of x³ is 3x², and since y² is treated as a constant multiplier, we get ∂f/∂x = 3x²y². To find ∂f/∂y, we treat x as a constant and differentiate x³y² with respect to y. The derivative of y² is 2y, and x³ is treated as a constant multiplier, giving us ∂f/∂y = 2x³y. Therefore, the gradient is ∇f = (3x²y², 2x³y).

We start by finding the gradient of the function f(x, y) = x² + y². The partial derivative with respect to x is 2x, and the partial derivative with respect to y is 2y, so the gradient is <2x, 2y>. Next, we evaluate the gradient at the point (1, 1), which gives us the vector <2, 2>. To find the directional derivative, we calculate the dot product of the gradient vector <2, 2> and the given unit direction vector <0.6, 0.8>. Performing the dot product gives us (2 * 0.6) + (2 * 0.8), which equals 1.2 + 1.6. Summing these values results in 2.8.

The directional derivative is given by D_u f = ||∇f|| cos(θ). This value is maximized when cos(θ) = 1, which occurs when the direction u is the same as the direction of the gradient (θ = 0). In this case, the directional derivative becomes ||∇f|| * 1 = ||∇f||. Thus, the maximum rate of change is the magnitude of the gradient itself.

To compute the gradient, we determine the partial derivatives with respect to each variable. First, we differentiate f(x, y) = x * e^y with respect to x while treating y as a constant. The derivative of x is 1, and e^y acts as a coefficient, so the result is e^y. Next, we differentiate with respect to y while treating x as a constant. The derivative of e^y is e^y, and x acts as a coefficient, so the result is x * e^y. Assembling these components into a vector, we obtain the gradient <e^y, x * e^y>.^y,>

To determine if the function increases or decreases, we look at the directional derivative in the direction of movement. The direction is North, represented by the unit vector j = <0, 1>. The directional derivative is the dot product of the gradient and the direction: ∇f · j = <0.5, -0.5> · <0, 1> = (0.5)(0) + (-0.5)(1) = -0.5. Since the result is negative, the altitude is decreasing, meaning you are descending.

First, we compute the gradient of the function f(x, y) = xy + y². The partial derivative with respect to x is y, and the partial derivative with respect to y is x + 2y. Evaluating this at the point (2, 1), we get a gradient of <1, 2 + 2(1)> which simplifies to <1, 4>. Next, we must normalize the direction vector v = <3, 4> because the formula requires a unit vector. The magnitude of v is √(3² + 4²) = 5. Dividing v by its magnitude, we get the unit vector u = <3/5, 4/5> or <0.6, 0.8>. Finally, we take the dot product of the gradient <1, 4> and the unit vector <0.6, 0.8>. This yields (1 * 0.6) + (4 * 0.8) = 0.6 + 3.2, which equals 3.8.

We interpret the significance of the gradient to find the maximum rate of increase. The direction of maximum increase is given by the gradient vector, but the *rate* (magnitude) of that increase is the magnitude (length) of the gradient vector. First, we find the gradient. The partial derivative with respect to x is 2x, and with respect to y is -3. At the point (2, 1), the gradient is <2(2), -3> = <4, -3>. To find the maximum rate, we calculate the magnitude of this vector: √(4² + (-3)²). This simplifies to √(16 + 9) = √(25), which equals 5.

We start by finding the gradient vector. The partial derivative with respect to x is cos(x), and with respect to y is -sin(y). Evaluating this at (0, 𝜋/2), we get cos(0) = 1 and -sin(𝜋/2) = -1, so the gradient is <1, -1>. The direction of steepest descent is the opposite of the gradient direction. Therefore, we multiply the gradient by -1 to get -<1, -1>, which results in the vector <-1, 1>.

The rate of increase in the direction of steepest increase is the magnitude of the gradient. We get that ∇f = (2x, 4y). At (1,1), ∇f = (2, 4). Its magnitude is √(2² + 4²) = √(4+16) = √20.

We use the alternative formula for the directional derivative: D_u f = ||gradient|| * cos(theta), where theta is the angle between the gradient and the direction vector. First, we calculate the magnitude of the gradient <2, 2>, which is √(2² + 2²) = √(8). We are given that the directional derivative is also √(8). Setting up the equation, we have √(8) = √(8) * cos(theta). Dividing both sides by √(8) gives cos(theta) = 1. The angle whose cosine is 1 is 0 degrees. This implies the direction u is exactly aligned with the gradient.

We extend the definition of the gradient to three variables by computing partial derivatives for x, y, and z. The partial derivative with respect to x is 2x. The partial derivative with respect to y is 2y. The partial derivative with respect to z is 2z. This gives the gradient vector field <2x, 2y, 2z>. We then substitute the coordinates of the point (1, 2, 3) into this vector. This yields <2(1), 2(2), 2(3)>, which simplifies to <2, 4, 6>.

We utilize the geometric property that the gradient vector is perpendicular (normal) to the level curve. The gradient is given as <4, -1>. The slope of this gradient vector is change-in-y over change-in-x, which is -1/4. Since the tangent line to the level curve is perpendicular to the gradient, its slope must be the negative reciprocal of the gradient's slope. The negative reciprocal of -1/4 is 4. Therefore, the slope of the tangent line to the level curve is 4.

We analyze the relationship between the gradient and the directional derivatives. If the directional derivative is 0 in a certain direction, the gradient must be perpendicular to that direction. Here, the gradient is perpendicular to <1, 1> and also perpendicular to <1, -1>. In a two-dimensional plane, the only vector that is perpendicular to two non-parallel vectors simultaneously is the zero vector. Therefore, the gradient vector must be <0, 0>, indicating a critical point (like a local maximum, minimum, or saddle point) or a flat region.