Using Properties of the GradientIn Exercises 29–38, find the gradient of the function and the maximum value of the directional derivative at the given point.
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Chapter 13 Solutions
Multivariable Calculus
- Using gradient rules Use the gradient rules of Exercise 85 to find the gradient of the following functions.arrow_forward(a) Find the gradient of f. (b) Evaluate the gradient at the point P. (c) Find the rate of change of f at P in the direction of the vector u.arrow_forwardUsing a Function (a) find the gradient of the function at P, (b) find a unit normal vector to the level curve f (x, y) = c at P, (c) find the tangent line to the level curve f (x, y) = c at P, and (d) sketch the level curve, the unit normal vector, and the tangent line in the xy-plane. f(x, y) = 9x2 + 4y2, c = 40, P(2, −1)arrow_forward
- f(x,y,z)=5sin(xyz), (a,b,c)=(1,1,π/4), and v⃗ =(5,−2–√,5). Calculate the directional derivative of f at the point (a,b,c)=(1,1,π/4) in the direction defined by v⃗ . Find the direction at (a,b,c)=(1,1,π/4) in which the rate of change of f is greatest. Find the maximum rate of change. Fill in the blank: f decreases the most at (a,b,c)=(1,1,π/4) in the direction ofarrow_forwardUsing Properties of the Derivative In Exercise 26, use the properties of the derivative to find the following. (a) r′(t) (b) d dt [u(t) − 2r(t)] (c) d dt [(3t)r(t)] (d) d dt [r(t) ∙ u(t)] (e) d dt [r(t) × u(t)] (f) d dt [u(2t)] 26. r(t) = sin ti + cos tj + tk, u(t) = sin ti + cos tj + 1 t k * only d ,e, f *arrow_forwardFind the gradient of the function f(x,y,z)=y^2ln(xz), at the point (1,2,e) ∇f(1,2,e)=arrow_forward
- a) find the rate of change in f(x,y) when moving in the direction of the vector (1,3) b) find the directional derivative of f(x,y) when moving in the direction of maximum increase c) find the direction of maximum decrease in f(x,y)arrow_forwardUsing a Function (a) find the gradient of the function at P, (b) find a unit normal vector to the level curve f (x, y) = c at P, (c) find the tangent line to the level curve f (x, y) = c at P, and (d) sketch the level curve, the unit normal vector, and the tangent line in the xy-plane. f(x, y) = 4x2 − y , c = 6, P(2, 10)arrow_forward1) Define the gradient vector, and explain what it means. Give an example of a real-world situation where knowing the gradient vector of a function or surface would be useful or important.arrow_forward
- Find the maximum rate of change of f at the given point and the direction in which it occurs. f(x, y, z) = x^2+y^2+z^2, (9,2,-8) maximum rate of change direction vectorarrow_forwardUsing a Function, (a) find the gradient of the function at P, (b) find a unit normal vector to the level curve f(x, y) = c at P, (c) find the tangent line to the level curve f(x, y) = c at P, and (d) sketch the level curve, the unit normal vector, and the tangent line in the xy-plane. f(x, y) = 9x2 − 4y2, c = 65, P(3, 2)arrow_forwardConsider the function f and the point P. f(x, y) = xey + yex, P(0, 0) ) Find the gradient of f and evaluate the gradient at the point P. ∇f(0, 0) Find the directional derivative of f at P in the direction of the vector v = (5, 1). Duf(0, 0) Find the maximum rate of change of f at P and the direction in which it occurs. maximum rate of change direction vectorarrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage