Chapter 6: Q1P (page 294)

Find the gradient of $w = x^{2} y^{3} z$ at $(1, 2, - 1)$ .

Short Answer

Expert verified

The gradient of function $w = x^{2} y^{3} z$ at $(1, 2, - 1)$ is $- 16 i - 12 j + 8 k$ .

Step by step solution

Given Information.

The given function is $w = x^{2} y^{3} z$ and reference point is $(1, 2, - 1)$ .

Definition of Gradient.

Gradient is a differential operator which is applied to vector function in three dimensions in order to give a vector whose three components are the partial derivatives of given function with respect to its three variables. It is represented by $\nabla$ .

Use the formula.

The formula states the expression mentioned below.

$\begin{array}{l} \nabla w = g r a d w \\ \nabla w = i \frac{\partial w}{\partial x} + y \frac{\partial w}{\partial y} + k \frac{\partial w}{\partial z} \end{array}$

Calculate the gradient.

Use the gradient formula for the given vector $w = x^{2} y^{3} z$ .

$\begin{array}{l} \nabla w = i \frac{\partial}{\partial x} (x^{2} y^{3} z) + j \frac{\partial}{\partial y} (x^{2} y^{3} z) + k \frac{\partial}{\partial z} (x^{2} y^{3} z) \\ \nabla w = i (2 x y^{3} z) + j (3 x^{2} y^{2} z) + k (x^{2} y^{3}) \end{array}$

Use the reference point given below.

$\begin{array}{l} x = 1, \\ y = 2, \\ z = - 1 \end{array}$

Now put the values in the equation.

$\begin{array}{l} \nabla w_{(1, 2, - 1)} = 2 (1) {(2)}^{3} (- 1) i + 3 {(1)}^{2} {(2)}^{2} (- 1) j + {(1)}^{2} {(2)}^{3} k \\ \nabla w_{(1, 2, - 1)} = - 16 i - 12 j + 8 k \end{array}$

The gradient of function $w = x^{2} y^{3} z$ at $(1, 2, - 1)$ is $- 16 i - 12 j + 8 k$ .

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Find the gradient of $w = x^{2} y^{3} z$ at $(1, 2, - 1)$ .

Short Answer

Step by step solution

Given Information.

Definition of Gradient.

Use the formula.

Calculate the gradient.

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Find the gradient of w=x2y3z at (1,2,-1).

Short Answer

Step by step solution

Given Information.

Definition of Gradient.

Use the formula.

Calculate the gradient.

One App. One Place for Learning.

Most popular questions from this chapter

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Find the gradient of $w = x^{2} y^{3} z$ at $(1, 2, - 1)$ .