Chapter 6: Q10P (page 295)

If the temperature in the $(x, y)$ plane is given by $T = x y - x$ , sketch a few isothermal curves, say for $T = 0, 1, 2, - 1, - 2$ . Find the direction in which the temperature changes most rapidly with distance from the point $(1, 1)$ , and the maximum rate of change. Find the directional derivative of T at $(1, 1)$ in the direction of the vector $3 i - 4 j$ Heat flows in the direction $螖 T$ (perpendicular to the isothermals). Sketch a few curves along which heat would flow.

Short Answer

Expert verified

Directional derivative is $\frac{- 4}{5}$ .

The direction in which the temperature changes most rapidly, $螖 T = j$ .

The maximum rate of change is $1$ .

Step by step solution

Given Information

The equation of the plane is $T = x y - x$ , vector $3 i - 4 j$ ,and point $(1, 1)$ .

Definition of Directional derivative.

The directional derivative is the rate of change along a unit vector.

The formula of the directional derivative $\frac{d 蠒}{d u}$ .

Find the Gradient.

Find equation of gradient.

Formula states the equation mentioned below.

$\begin{array}{l} 螖 T = \frac{鈭� f}{鈭� x} i + \frac{鈭� f}{鈭� y} j + \frac{鈭� f}{鈭� z} k = 0 \\ 螖 T = (y_{0} - 1) i - j = 0 \end{array}$

Put $(1, 1)$ in the above equation.

$螖 T = j$

The direction in which the temperature changes most rapidly is $螖 T = j$ .

The maximum rate of change is $|螖 T| = |j|$ .

$|螖 T| = 1$

Hence the maximum rate of change is $1$ .

Find the Directional derivative.

Find directional derivative.

Put the given value in the formula.

$\begin{matrix} 蠒 = (0, 1, 0) \\ u = \frac{1}{5} (3, 4, 0) \end{matrix}$

The equation becomes as mentioned below.

$\begin{matrix} \frac{d 蠒}{d u} = (0, 1, 0) 脳 \frac{1}{5} (3, 4, 0) \\ \frac{d 蠒}{d u} = \frac{- 4}{5} \end{matrix}$

Hence the directional derivative is $\frac{- 4}{5}$ .

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Short Answer

Step by step solution

Given Information

Definition of Directional derivative.

Find the Gradient.

Find the Directional derivative.

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