Calculate the directional derivative of&nbsp;f(x,y,z)=xy+yz+zx&nbsp;at P(1, -1, 3) in the direction of Q(2, 4, 5).

We should find the directional derivative of the function f(x,y,z)=xy+yz+zx at the point P(1,−1,3) in the direction of the point Q(2,4,5) The partial derivatives are fx(x,y,z)=y+z, fy(x,y,z)=x+z, fz(x,y,z)=x+y The vector is given as v=PQ→=&lt;1−2,−1−4,3−5≥&lt;1,5,2&gt; The unit vector in the direction of v, which we will denote by u, is: u=&lt;112+52+22,512+52+22,212+52+22≥&lt;130,530,230&gt; Now by formula 9 from the textbook (Duf(x,y,z)=△f(x,y,z)⋅u), we have: Duf(1,−1,3)=&lt;2,4,0&gt;⋅&lt;130,530,230≥ =2230

"Calculate the directional derivative off(x,y,z)=xy+yz+zxat P(1, -1, 3)..." was answered by our community

vakirnarhh

Answered question

2021-11-15

Calculate the directional derivative of $f (x, y, z) = x y + y z + z x$ at P(1, -1, 3) in the direction of Q(2, 4, 5).

Answer & Explanation

Charles Wee

Beginner2021-11-16Added 14 answers

We should find the directional derivative of the function $f (x, y, z) = x y + y z + z x$ at the point $P (1, - 1, 3)$ in the direction of the point $Q (2, 4, 5)$
The partial derivatives are
$f_{x} (x, y, z) = y + z, f_{y} (x, y, z) = x + z, f_{z} (x, y, z) = x + y$
The vector is given as $v = \vec{P Q} = < 1 - 2, - 1 - 4, 3 - 5 \geq < 1, 5, 2 >$
The unit vector in the direction of v, which we will denote by u, is:
$u = < \frac{1}{\sqrt{1^{2} + 5^{2} + 2^{2}}}, \frac{5}{\sqrt{1^{2} + 5^{2} + 2^{2}}}, \frac{2}{\sqrt{1^{2} + 5^{2} + 2^{2}}} \geq < \frac{1}{\sqrt{30}}, \frac{5}{\sqrt{30}}, \frac{2}{\sqrt{30}} >$
Now by formula 9 from the textbook $(D_{u} f (x, y, z) = △ f (x, y, z) \cdot u)$ , we have:
$D_{u} f (1, - 1, 3) = < 2, 4, 0 > \cdot < \frac{1}{\sqrt{30}}, \frac{5}{\sqrt{30}}, \frac{2}{\sqrt{30}} \geq$
$= \frac{22}{\sqrt{30}}$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?