The maximum value of the directional derivative of the function ϕ = 2x^{2} + 3y^{2} + 5z^{2} at point (1, 1, -1) is

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BPSC Asstt. Prof. ME Held on Nov 2015 (Advt. 22/2014)

Option 3 : \(\sqrt{152}\)

CT 1: Ratio and Proportion

3742

10 Questions
16 Marks
30 Mins

**Concept:**

**Gradient:**

For a scalar function ϕ(x, y, z), the gradient is the maximum rate of change which is given by-

**\(grad\;ϕ=∇ϕ\Rightarrow\left(\hat i\frac{\partial}{\partial x}\;+\;\hat j\frac{\partial}{\partial y}\;+\;\hat k\frac{\partial}{\partial z} \right)ϕ\)**

**Directional derivative:**

It gives the rate of change of scalar point function in a particular direction. The maximum magnitude of the directional derivative is the magnitude of the gradient.

**Calculation:**

**Given:**

ϕ = 2x2 + 3y2 + 5z2

\(grad\;ϕ=∇ϕ\Rightarrow\left(\hat i\frac{\partial}{\partial x}\;+\;\hat j\frac{\partial}{\partial y}\;+\;\hat k\frac{\partial}{\partial z} \right)ϕ\)

\(∇ϕ=\left(\hat i\frac{\partial}{\partial x}\;+\;\hat j\frac{\partial}{\partial y}\;+\;\hat k\frac{\partial}{\partial z} \right){(2x^2\;+\;3y^2\;+\;5z^2)}\)

∇ϕ = 4xî + 6yĵ + 10zk̂

∇ϕ_{(1, 1, -1)} = 4î + 6ŷ - 10k̂

Maximum value is given by the magnitude,

\(\therefore\;\sqrt{4^2\;+\;6^2\;+\;(-10)^2}\Rightarrow\sqrt{152}\)