Question

Suppose that the temperature at the point \(x,y,x)\ is given by the formula \[W(x,y,z)=100-x^2-y^2-z^2\]

The units of distance in space are in meters

a)find the rate of change of temperature at the point (3, -4, 5) if we start moving at the speed of one meter per second on the line of symmetric equations \[\frac{x-3}{3}=\frac{y+4}{4}=z-5\]

b) In what direction does W increase most rapidly at the point (3,-4,5)?

c)what is the maximal value of the directional derivative at (3,-4,5)

Answer 1

so I think I need to find \(\nabla W\) ?

Answer 2

Suppose that the temperature at the point \(x,y,x\) is given by the formula \[W(x,y,z)=100-x^2-y^2-z^2\]
The units of distance in space are in meters

a)find the rate of change of temperature at the point (3, -4, 5) if we start moving at the speed of one meter per second on the line of symmetric equations \[\frac{x-3}{3}=\frac{y+4}{4}=z-5\]
b) In what direction does W increase most rapidly at the point (3,-4,5)?

c)what is the maximal value of the directional derivative at (3,-4,5)

Answer 3

So the direction vector of the line is \((3,4,1)\), right? Then the directional derivative of W in that direction is simply the dot product of the gradient of W and the normalized (unit) above vector, times the speed (one).

The direction of maximal increase is always the gradient. Refer to your text.

C is very similar to B. (It's the same).

Answer 4

how did you get the direction vector of the line?

Answer 5

that's what I couldn't figure out. The answer says it's \(\vec{v}=(3,4,1)\) as well. With no explanation.

Answer 6

I had to look at some old calculus books... D:

So you have the general vector form of a line, right?\[\mathbf{r}=\mathbf{r}_0+\mathbf{v}t\]So this is just a series of parametric equations. In \(\mathbb{R}^3\):\[x=x_0+v_x t\]\[y=y_0+v_y t\]\[z=z_0+v_z t\]And after a little manipulation:\[\frac{x-x_0}{v_x}=\frac{y-y_0}{v_y}=\frac{z-z_0}{v_z}\]From which calculation of \(\mathbf{v}\) is simply identifying the reciprocal of the coefficient on each variable.

I think my "L" key is stuck. It makes typing so difficut.

Answer 7

Oh sweet thanks so much! Turns out he gave a shortcut by putting it in that form but I would have gotten 0 marks out of 10 for not identifying that!