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Created this as a copy of Find directional derivative of a scalar field.

Name	Status	Author	Last Modified
Find directional derivative of a scalar field	Ready to use	Newcastle University Mathematics and Statistics	31/01/2024 14:58
Harry's copy of Find directional derivative of a scalar field,	draft	Harry Flynn	09/03/2018 14:28
Simon's copy of Find directional derivative of a scalar field,	draft	Simon Thomas	03/04/2019 10:12
Ugur's copy of Find directional derivative of a scalar field,	Ready to use	Ugur Efem	07/06/2023 16:47

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Question statement

You are given the scalar field $f=\simplify{{a1}*x^{p1}*y^{p2}*z^{p3}+{b1}*x^{p4}*y^{p5}*z^{p6}}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a1	number	-7
b1	number	8
gradfq	vector	vector(39,-14,-64)
lenu	number	7.3484692283
p1	integer	1
p2	integer	1
p3	integer	0
p4	integer	2
p5	integer	0
p6	integer	2
q	vector	vector(2,-1,-1)
u	vector	vector(3,-3,6)
uhat	vector	Vector with 3 components
uhatdotgradfq	number	-30.619

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Generated value: `number`

-7

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Ask the student a question, and give any hints about how they should answer this part.

Calculate $\boldsymbol{\nabla}f$ .

$\boldsymbol{\nabla}f=($ Gap 0. $,$ Gap 1. $,$ Gap 2. $)$ .

Advice

Note that in this advice, the full calculator display is used in the calculation of each step; any rounding is purely for display clarity.

The gradient of $f$ is given by

$\begin{align}\boldsymbol{\nabla}f&=\pmatrix{\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}}\\&=\pmatrix{\simplify{{p1*a1}*x^{p1-1}*y^{p2}*z^{p3}+{p4*b1}*x^{p4-1}*y^{p5}*z^{p6}},\simplify{{p2*a1}*x^{p1}*y^{p2-1}*z^{p3}+{p5*b1}*x^{p4}*y^{p5-1}*z^{p6}},\simplify{{p3*a1}*x^{p1}*y^{p2}*z^{p3-1}+{p6*b1}*x^{p4}*y^{p5}*z^{p6-1}}},\end{align}$

by straight forward partial differentiation.

The gradient of $f$ at the point $\boldsymbol{q}=\pmatrix{\var{q[0]},\var{q[1]},\var{q[2]}}$ is found by substituting $\boldsymbol{q}$ into $\boldsymbol{\nabla}f$ , hence

$\boldsymbol{\nabla}f\pmatrix{\var{q[0]},\var{q[1]},\var{q[2]}}=\pmatrix{\var{gradfq[0]},\var{gradfq[1]},\var{gradfq[2]}}.$

The unit vector $\boldsymbol{\hat{u}}$ in the direction of $\boldsymbol{u}$ is given by

$\begin{align}\boldsymbol{\hat{u}}=\frac{\boldsymbol{u}}{\lvert\boldsymbol{u}\rvert}&=\frac{1}{\sqrt{(\var{u[0]})^2+(\var{u[1]})^2+(\var{u[2]})^2}}\pmatrix{\var{u[0]},\var{u[1]},\var{u[2]}}\\&=\frac{1}{\var{precround(lenu,3)}}\pmatrix{\var{u[0]},\var{u[1]},\var{u[2]}}\\&=\pmatrix{\var{precround(uhat[0],3)},\var{precround(uhat[1],3)},\var{precround(uhat[2],3)}}.\end{align}$

The directional derivative $\frac{\partial f}{\partial\boldsymbol{u}}$ , of the scalar field $f$ , in the direction of $\boldsymbol{u}$ , at the point $\boldsymbol{q}$ is given by

$\begin{align}\frac{\partial f}{\partial\boldsymbol{u}}\pmatrix{\var{u[0]},\var{u[1]},\var{u[2]}}&=\boldsymbol{\hat{u}\cdot\nabla}f\pmatrix{\var{q[0]},\var{q[1]},\var{q[2]}}\\&=\pmatrix{\var{precround(uhat[0],3)},\var{precround(uhat[1],3)},\var{precround(uhat[2],3)}}\boldsymbol{\cdot}\pmatrix{\var{gradfq[0]},\var{gradfq[1]},\var{gradfq[2]}}\\&=\var{uhatdotgradfq}\;\text{to 3d.p., using the full calculator display for the answers in the previous part.}\end{align}$

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This question is used in the following exam:

Vector Calculus by Ugur Efem in Ugur's workspace.

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Ugur's copy of Find directional derivative of a scalar field,

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Ugur's copy of Find directional derivative of a scalar field,

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