# Temperature as a function of 3 variables...

1. Apr 9, 2016

### TheSodesa

Mod note: moved from non-HW forum section
Apparently doing math homework on Physics Forums helps me think more clearly. The d-part of this question gave me a headache, and I already wrote all of this LaTeX code on the homework side of this site, so I figured I'd post it even though I solved it.

Parts a-c might have small mistakes as well, but duck it, I'm done with this question.

1. The problem statement, all variables and given/known data

Let $T(x,y,z) = 10+xy+xz+yz$. A falcon is at the point $P_0 = (x_0, y_0, z_0) = (3,2,1)$.

a) What is the temperature at the point $P_0$.
b) What is the instantaneous rate of change of temperature experienced by the falcon as it flies towards the point $P_1 = (2,4,4)$
c) In what direction should the falcon fly in if it wanted to experience the largest (positive) rate of change in temperature?
d) Give a direction in which the temperature does not change (there are infinitely many).

2. Relevant equations

The Gradient: $$\nabla T(P)= (\frac{\partial T}{\partial x},\frac{\partial T}{\partial y},\frac{\partial T}{\partial z})$$

Directional derivative: $$D_u T(P) = \nabla T(P) \cdot \hat{u}$$

3. The attempt at a solution

To start with, $\nabla T(P_0)= (\frac{\partial T}{\partial x_0},\frac{\partial T}{\partial y_0},\frac{\partial T}{\partial z_0}) = (y_0+z_0, x_0+z_0, x_0+y_0) = (3, 4, 5)$

a) Simply plug in the values of x,y and z to get $T(P_0) = 21$

b) Here
$$\vec{u} = P_1 - P_0 = (2,4,4) - (3,2,1) = (-1,2,3)$$
and
$$\hat{u} = \frac{\vec{u}}{||\vec{u}||} = \frac{(-1, 2, 3)}{\sqrt{1^2 + 2^2 + 3^3}} = \frac{(-1, 2, 3)}{ \sqrt{14} }$$

Therefore the directional derivative:

$$D_u T(P_0) = \nabla T(P_0) \cdot \hat{u} = (3, 4, 5) \cdot \frac{(-1, 2, 3) }{ \sqrt{14} } = \frac{20}{\sqrt{14}}$$

c) In the direction of the gradient.

Here $\hat{u} = \frac{\nabla T(P_0)}{||\nabla T(P_0)||} = \frac{1}{5 \sqrt{2}}(3,4,5)$.

Therefore

$$D_u T(P_0) = \nabla T(P_0) \cdot \hat{u} = (3,4,5) \cdot (\frac{3}{5 \sqrt{2}},\frac{4}{5 \sqrt{2}},\frac{5}{5 \sqrt{2}}) = \frac{9}{5 \sqrt{2}} + \frac{16}{5 \sqrt{2}} + \frac{25}{5 \sqrt{2}} =\frac{10}{\sqrt{2}}$$

d) This is where I ran into trouble. I knew that in the direction of the level curve (if such a thing exists in 4D), the directional derivative is equal to 0. Using this fact and the fact that $||\hat{u}|| = 1$, I get two seemingly messy equations, that I was initially unable to solve:

If we mark $\hat{u} = (x,y,z)$
$$D_u T(P_0) = \nabla T(P_0) \cdot \hat{u} = 3x+4y+5z = 0$$
and
$$\sqrt{x^2+y^2+z^2} = 1 \iff x^2 + y^2 + z^2 = 1$$
More clearly:

\begin{cases}
3x+4y+5z = 0 \ || \ solve \ for \ z\\
x^2 + y^2 + z^2 = 1
\end{cases}

\begin{cases}
z = \frac{1}{5}(-3x - 4y) \\
x^2 + y^2 + z^2 = 1 \ || \ Substitute \ z \ here
\end{cases}

\begin{cases}
z = \frac{1}{5}(-3x - 4y) \\
x^2 + y^2 + (\frac{1}{5}(-3x - 4y))^2 = 1 \ || \ Multiply\ (-3x - 4y))^2
\end{cases}

\begin{cases}
z = \frac{1}{5}(-3x - 4y) \\
x^2 + y^2 + \frac{1}{25}(9x^2 + 24xy + 16y^2) = 1 \ || \ \cdot 25
\end{cases}

\begin{cases}
z = \frac{1}{5}(-3x - 4y) \\
25x^2 + 25y^2 + (9x^2 + 24xy + 16y^2) = 25
\end{cases}

\begin{cases}
z = \frac{1}{5}(-3x - 4y) \\
34x^2 + 24xy + 41y^2 = 25 \iff 34x^2 + 24xy + 41y^2 -25 = 0 \ || \ Solve \ for \ x
\end{cases}

$$x = \frac {-24y \pm \sqrt{(24y)^2 - 4(34)(41y^2 - 25)}}{2(34)}$$
$$x = \frac {-24y \pm \sqrt{-200(25y^2-17)}}{68}$$

Now to make things simple, let the discriminant be equal to zero:
$$-200(25y^2-17)=0 \iff 5000y^2 = 3400 \iff y= \pm \sqrt{\frac{3400}{5000}} = \pm \frac{\sqrt{17}}{5}$$

Again, to make things simple, plug positive y into x:
\begin{array}{ll}
x &= \frac {-24y \pm \sqrt{-200(25y^2-17)}}{68}\\
x &= \frac {-24(\frac{\sqrt{17}}{5}) \pm \sqrt{0}}{68} \ || \ Remember \ the \ discriminant \ was \ zero \\
x &= \frac {-24(\frac{\sqrt{17}}{5})}{68} = \frac{-6}{5 \sqrt{17}}
\end{array}

Then by pugging x and y into z:
\begin{array}{ll}
z &= \frac{1}{5}(-3x - 4y)\\
&= \frac{1}{5}(-3({\frac{-6}{5 \sqrt{17}}) - 4(\frac{\sqrt{17}}{5}})\\
&= \frac{-2}{\sqrt{17}}
\end{array}

Therefore
\begin{cases}
x = \frac{-6}{5 \sqrt{17}}\\
y = \frac{\sqrt{17}}{5}\\
z = \frac{-2}{\sqrt{17}}
\end{cases}
and the direction we want to head in is

$$\hat{u} = (x,y,z) = (\frac{-6}{5 \sqrt{17}},\frac{\sqrt{17}}{5},\frac{-2}{\sqrt{17}})$$

Last edited by a moderator: Apr 9, 2016
2. Apr 9, 2016