Question

Suppose that over a certain region of space the electrical potential \(V\) is given by \(V(x,y,z) = 5{x^2} - 3xy + xyz\).

(a) Find the rate of change of the potential at \(P(3,4,5)\) in the direction of the vector \({\bf{v}} = {\bf{i}} + {\bf{j}} - {\bf{k}}\).

(b) In which direction does \(V\) change most rapidly at \(P\) ?

(c) What is the maximum rate of change at \(P\) ?

Short answer

a) The rate of change of \(V(x,y,z) = 5{x^2} - 3xy + xyz\) at the point \(P(3,4,5)\) in the direction of \({\rm{v}} = {\rm{i}} + {\rm{j}} - {\rm{k}}\) is, \({D_u}V(3,4,5) = \frac{{32}}{{\sqrt 3 }}\).

b) \(V\) increases fastest in the direction of \(\nabla V(3,4,5) = \langle 38,6,12\rangle \).

c) The maximum rate of change of \(V(x,y,z) = 5{x^2} - 3xy + xyz\) at the point \(P(3,4,5)\) is \(|\nabla V(3,4,5)| = 2\sqrt {406} \).

Step-by-step solution

The rate of change of vector and maximum value of the directional derivative.

Suppose\(f\)is a differentiable function of two variables. The rate of change is defined as\({D_u}f(x,y) = \nabla f(x,y) \cdot u\)where\(u\)is unit vector.

Suppose\(f\)is a differentiable function of two or three variables. The maximum value of the directional derivative\({D_u}f(x)\)is\(|\nabla f(x)|\)and it occurs when u has the same direction as the gradient vector\(\nabla f(x){\rm{ }}\)

The rate of change of \(V(x,y,z) = 5{x^2} - 3xy + xyz\) at the point \(P(3,4,5)\).

a)

The given temperature function is, \(V(x,y,z) = 5{x^2} - 3xy + xyz\).

The direction of the vector \(v = i + j - k\).

The unit vector \(u\) is computed as follows,

\(\begin{aligned}{l}u = \frac{1}{{\sqrt {{1^2} + {1^2} + {{( - 1)}^2}} }}\langle 1,1, - 1\rangle \\ = \frac{1}{{\sqrt {1 + 1 + 1} }}\langle 1,1, - 1\rangle \end{aligned}\)

\( = \frac{1}{{\sqrt 3 }}\langle 1,1, - 1\rangle \)

Thus, the unit vector \(u\) is \(\frac{1}{{\sqrt 3 }}\langle 1,1, - 1\rangle \).

Then the gradient of \(V(x,y,z)\) is,

\(\begin{aligned}{l}\nabla V(x,y,z) = \left\langle {{V_x},{V_y},{V_z}} \right\rangle \\ = \left\langle {\frac{\partial }{{\partial x}}\left( {5{x^2} - 3xy + xyz} \right),\frac{\partial }{{\partial y}}\left( {5{x^2} - 3xy + xyz} \right),\frac{\partial }{{\partial z}}\left( {5{x^2} - 3xy + xyz} \right)} \right\rangle \\ = \langle (10x - 3y + yz),(0 - 3x + xz),(0 - 0 + xy)\rangle \\ = \langle (10x - 3y + yz),( - 3x + xz),(xy)\rangle \end{aligned}\)

Thus, the gradient of \(V(x,y,z)\) is \(\nabla V(x,y,z) = \langle (10x - 3y + yz),( - 3x + xz),(xy)\rangle \).

The rate of change, \({D_u}V(3,4,5) = \nabla V(3,4,5) \cdot {\rm{u}}\) is calculated as follows.

\(\begin{aligned}{l}{D_u}T(3,4,5) = \langle (10x - 3y + yz),( - 3x + xz),(xy)\rangle \cdot \frac{1}{{\sqrt 3 }}\langle 1,1, - 1\rangle \\ = \langle (10(3) - 3(4) + (4)(5)),( - 3(3) + (3)(5)),((3)(4))\rangle \cdot \frac{1}{{\sqrt 3 }}\langle 1,1, - 1\rangle \\ = \langle 30 - 12 + 20,( - 9 + 15),(12)\rangle \cdot \frac{1}{{\sqrt 3 }}\langle 1,1, - 1\rangle \\ = \langle 38,6,12\rangle \cdot \frac{1}{{\sqrt 3 }}\langle 1,1, - 1\rangle \end{aligned}\)

Simplify further as,

\(\begin{aligned}{l}{D_u}V(3,4,5) = \langle 38,6,12\rangle \cdot \frac{1}{{\sqrt 3 }}\langle 1,1, - 1\rangle \\ = \frac{{38}}{{\sqrt 3 }} + \frac{6}{{\sqrt 3 }} - \frac{{12}}{{\sqrt 3 }}\\ = \frac{{38 + 6 - 12}}{{\sqrt 3 }}\\ = \frac{{32}}{{\sqrt 3 }}\end{aligned}\)

The rate of change of \(V(x,y,z) = 5{x^2} - 3xy + xyz\) at the point \(P(3,4,5)\) in the direction of \({\rm{v}} = {\rm{i}} + {\rm{j}} - {\rm{k}}\) is, \({D_u}V(3,4,5) = \frac{{32}}{{\sqrt 3 }}\).

The direction in which the potential function increases fastest at \(P(3,4,5)\).

b)

From part (a), the gradient of \(V(x,y,z)\) is \(\nabla V(x,y,z) = \langle (10x - 3y + yz),( - 3x + xz),(xy)\rangle \).

Thus, V increases fastest along its gradient vector.

The gradient vector of the temperature \(V(x,y,z)\) at the point \(P(3,4,5)\) is computed as follows,

\(\begin{aligned}{l}\nabla V(x,y,z) = \langle (10x - 3y + yz),( - 3x + xz),(xy)\rangle \nabla T(3,4,5)\\ = \langle (10(3) - 3(4) + (4)(5)),( - 3(3) + (3)(5)),((3)(4))\rangle \\ = \langle 30 - 12 + 20,( - 9 + 15),(12)\rangle \\ = \langle 38,6,12\rangle \end{aligned}\)

Thus, the potential increases fastest along the gradient vector \(\nabla V(3,4,5) = \langle 38,6,12\rangle \).

The maximum rate of change of \(V(x,y,z) = 5{x^2} - 3xy + xyz\) at the point\(P(3,4,5)\).

(c)

The given function is, \(V(x,y,z) = 5{x^2} - 3xy + xyz\).

From part (b), the maximum rate of change occur in the direction of \(\nabla V(3,4,5) = \langle 38,6,12\rangle \).

Then the maximum rate \(|\nabla f(x)|\) is calculated as follows,

\(\begin{aligned}{l}|\nabla V(3,4,5)| = |\langle 38,6,12\rangle |\\ = \sqrt {{{38}^2} + {6^2} + {{12}^2}} \\ = \sqrt {1444 + 36 + 144} \\ = \sqrt {1624} \end{aligned}\)

Thus, the maximum rate is \(|\nabla V(3,4,5)| = 2\sqrt {406} \).