Question

Let \(f\) be a function of two variables that has continuous partial derivatives and consider the points \(A(1,3),B(3,3)\), \(C(1,7)\), and \(D(6,15)\). The directional derivative of \(f\) at \(A\) in the direction of the vector \(\overrightarrow {AB} \) is 3 and the directional derivative at \(A\) in the direction of \(\overrightarrow {AC} \) is 26 . Find the directional derivative of \(f\) at \(A\) in the direction of the vector \(\overrightarrow {AD} \).

Short answer

The directional derivative of the function \(f\) at the point \(A\) in the direction of the vector \(\overrightarrow {AD} \) is \(\frac{{327}}{{13}}\).

Step-by-step solution

The directional derivative of function.

"The directional derivative of the function\(\nabla f(x,y,z)\)at\(f\left( {{x_0},{y_0},{z_0}} \right)\)in the direction of unit vector\(u = \langle a,b,c\rangle \)is\({D_u}f(x,y,z) = \nabla f(x,y,z) \cdot u\), where

\(\begin{aligned}{l}\nabla f(x,y,z) = \left\langle {{f_x},{f_y},{f_z}} \right\rangle \\ = \frac{{\partial f}}{{\partial x}}i + \frac{{\partial f}}{{\partial y}}j + \frac{{\partial f}}{{\partial z}}k\end{aligned}\)."

Compute the unit vector in the direction \(\overrightarrow {AB} \).

Compute the unit vector in the direction \(\overrightarrow {AB} \) is as follows,

\(\begin{aligned}{l}\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} \\ = (3 - 1)i + (3 - 3)j\\ = 2i\end{aligned}\)

The unit vector \(u\),

\(\begin{aligned}{l}u = \frac{{\overrightarrow {AB} }}{{|\overrightarrow {AB} |}}\\ = \frac{{2i}}{{\sqrt {{2^2}} }}\\ = i\end{aligned}\)

Thus, the unit vector in the direction \(\overrightarrow {AB} \) is \(i\).

Compute the unit vector in the direction \(\overrightarrow {AC} \).

Compute the unit vector in the direction \(\overrightarrow {AC} \) is as follows,

\(\begin{aligned}{l}\overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} \\ = (1 - 1)i + (7 - 3)j\\ = 4j\end{aligned}\)

The unit vector \(u\),

\(\begin{aligned}{l}u = \frac{{AC}}{{|\overrightarrow {AC} |}}\\ = \frac{{4j}}{{\sqrt {{4^2}} }}\\ = j\end{aligned}\)

Thus, the unit vector in the direction \(\overrightarrow {AC} \) is \(j\).

Compute the unit vector in the direction \(\overrightarrow {AD} \).

Compute the unit vector in the direction \(\overrightarrow {AD} \) is as follows,

\(\begin{aligned}{l}\overrightarrow {AD} = \overrightarrow {OD} - \overrightarrow {OA} \\ = (6 - 1)i + (15 - 3)j\\ = 5i + 12j\end{aligned}\)

The unit vector \(u\),

\(\begin{aligned}{l}u = \frac{{\overrightarrow {AD} }}{{|\overrightarrow {AD} |}}\\ = \frac{{5i + 12j}}{{\sqrt {{5^2} + {{12}^2}} }}\\ = \frac{5}{{13}}i + \frac{{12}}{{13}}j\\ = \left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \end{aligned}\)

Thus, the unit vector in the direction \(\overrightarrow {AD} \) is \(\left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \).

Find the value of \({f_x}\) and \({f_y}\).

The value of \({f_x}(1,3)\) is computed as follows,

\(\begin{aligned}{l}{D_{\overrightarrow {AB} }}f(1,3) = \nabla f(1,3) \cdot u\\3 = \left\langle {{f_x}(1,3),{f_y}(1,3)} \right\rangle \cdot \langle 1,0\rangle \\{f_x}(1,3) = 3\end{aligned}\)

Thus, the value of \({f_x}(1,3)\) is 3 .

The value of \({f_y}(1,3)\) is computed as follows,

\(\begin{aligned}{l}{D_{\overrightarrow {AC} }}f(1,3) = \nabla f(1,3) \cdot u\\26 = \left\langle {{f_x}(1,3),{f_y}(1,3)} \right\rangle \cdot \langle 0,1\rangle \\{f_y}(1,3) = 26\end{aligned}\)

Thus, the value of \({f_y}(1,3)\) is 26 .

Find the directional derivative of the function \(f\) at the point \(A\).

Compute the directional derivative of the function \(f\) at the point \(A\) in the direction of the vector \(\overrightarrow {AD} \) is as follows,

\(\begin{aligned}{l}{D_{\overrightarrow {AD} }}f(1,3) = \nabla f(1,3) \cdot u\\ = \left\langle {{f_x}(1,3),{f_y}(1,3)} \right\rangle \cdot \left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \\ = \langle 3,26\rangle \cdot \left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \\ = \frac{{327}}{{13}}\end{aligned}\)

Thus, the required directional derivative is, \({D_{\overrightarrow {AD} }}f(1,3) = \frac{{327}}{{13}}\).