Solve the rate of change of f at P in the direction of the vector u.&nbsp;f(x,y,z)=y2exyz,&nbsp;P(0,1,−1),&nbsp;u=&lt;313,&nbsp;413,&nbsp;1213&gt;

Recall that: Rate of change of at P(a,b,c) in the direction of u is Duf Where Duf(a,b,c)=▽f(a,b,c)⋅u Where u is unit vector From part (b), we know that ▽f(0,1,−1)=&lt;−1,2,0&gt; Given that u=&lt;313,413,1213&gt;, Note that it is already a unit vector Duf(0,1,−1)=▽f(0,1,−1)⋅u Duf(0,1,−1)=&lt;−1,2,0&gt;⋅&lt;313,413,1213&gt; Duf(0,1,−1)=−313+813+0=513 Result: Duf(0,1,−1)=513

Question: "Solve the rate of change of f at..."

College
Algebra
Linear algebra
Vectors and spaces

Jaya Legge

Answered question

2021-09-13

Solve the rate of change of f at P in the direction of the vector u. $f (x, y, z) = y^{2} e^{x} y z, P (0, 1, - 1), u = < \frac{3}{13}, \frac{4}{13}, \frac{12}{13} >$

Answer & Explanation

opsadnojD

Skilled2021-09-14Added 95 answers

Recall that:
Rate of change of at $P (a, b, c)$ in the direction of u is $D_{u} f$
Where $D_{u} f (a, b, c) = ▽ f (a, b, c) \cdot u$
Where u is unit vector
From part (b), we know that
$▽ f (0, 1, - 1) = < - 1, 2, 0 >$
Given that $u = < \frac{3}{13}, \frac{4}{13}, \frac{12}{13} >$ , Note that it is already a unit vector
$D_{u} f (0, 1, - 1) = ▽ f (0, 1, - 1) \cdot u$
$D_{u} f (0, 1, - 1) = < - 1, 2, 0 > \cdot < \frac{3}{13}, \frac{4}{13}, \frac{12}{13} >$
$D_{u} f (0, 1, - 1) = - \frac{3}{13} + \frac{8}{13} + 0 = \frac{5}{13}$
Result: $D_{u} f (0, 1, - 1) = \frac{5}{13}$

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