Heat Transfer A metal plate is placed on the x y -plane in...

Heat Transfer A metal plate is placed on the $x y$ -plane in such a way that the temperature $T$ at any point $(x, y)$ is given by $T=e^{x}(\sin x+\sin y)^{\circ} \mathrm{C}$ (a) What is the rate of change in temperature at (0,0) in the direction of $3 \mathbf{i}-4 \mathbf{j} ?$ (b) At (0,0) , in what direction is the temperature increasing most rapidly? (c) In what direction is the temperature decreasing most rapidly? (d) In what direction does the temperature remain the same?

Heat Transfer A metal plate is placed on the $x y$ -plane in such a way that the temperature $T$ at any point $(x, y)$ is given by $T=e^{x}(\sin x+\sin y)^{\circ} \mathrm{C}$
(a) What is the rate of change in temperature at (0,0) in the direction of $3 \mathbf{i}-4 \mathbf{j} ?$
(b) At (0,0) , in what direction is the temperature increasing most rapidly?
(c) In what direction is the temperature decreasing most rapidly?
(d) In what direction does the temperature remain the same?

Key Concepts

Unit Vector Normalization

Unit vector normalization involves converting any given vector into a unit vector by dividing it by its magnitude. This process is essential when computing directional derivatives since the directional change must be assessed relative to a unit direction.

Gradient

The gradient of a scalar function is a vector that consists of its partial derivatives with respect to its independent variables. It points in the direction of the steepest ascent of the function and its magnitude represents the rate of maximum increase. In problems of heat transfer, for example, the gradient indicates the direction of greatest temperature increase on a plate.

Directional Derivative

The directional derivative represents the rate at which a function changes at a point in a specified direction. It is calculated as the dot product of the gradient vector and a unit vector in the desired direction. This concept is crucial to determine how temperature varies along any given direction on a surface.

Maximum Rate of Increase

The maximum rate of increase of a function at a given point is always in the direction of the gradient vector. The magnitude of the gradient quantifies this maximum rate. Physically, in heat transfer contexts, this indicates the direction in which the temperature increases most rapidly.

Maximum Rate of Decrease

Conversely, the maximum rate of decrease of a function occurs in the direction opposite to the gradient vector. This indicates the direction in which a function, such as temperature, decreases most rapidly.

Level Curves

Level curves are sets of points where the function takes on a constant value. The direction tangent to these curves corresponds to zero change in the function's value. Thus, moving along these curves results in no change in the quantity being measured, which is essential for understanding scenarios where the temperature remains the same.

Transcript

00:01 For this problem, we are told that a metal place is placed on the xy plane in such a way that the temperature t at any point xy is given by t equals e to power of x times sine of x plus sine of y degrees celsius.

00:12 In part a, we are asked, what is the rate of change in temperature at the point zero zero in the direction of 3i minus 4j? so this is essentially asking us for a directional derivative in part a.

00:22 First thing that we want to do is we want to figure out what our gradient of our function is going to be at the point zero.

00:28 0, and that is actually something that we'll be using in later problems.

00:33 But the gradient, first of all, i'll write down what we have in the x -axis.

00:37 So we'll need to apply product rule.

00:40 We'd have each the power of x times sine of x plus cos of x times e to the power of x.

00:49 And then in the y -axis, we would have e to the power of x times cos of y.

00:56 We want to evaluate this at the point 0 .0.

00:58 So the sign term will go to zero, so we'd be left with just one times east power of zero, so we'd have one, and then one as our gradient.

01:10 Then we want our directional derivative, which is going to be in the direction of 3i minus 4j.

01:17 So i'll write it in this ijk notation here, 3i minus 4j over the square root of 3 squared plus 4 squared, so 9 plus 16, which would be 20.

01:29 So we'd have that it's going to be 1 over 5 times 3 i had minus 4 j hat...

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