Question

Unit vector normal to the equipotential surface of 𝑉(𝑥, 𝑦, 𝑧) =4𝑥²+𝑦² + 𝑧 at (1,2,1) is given by (𝑎𝑖̂+ 𝑏𝑗̂+ 𝑐𝑘̂). The value of |𝑏| is (rounded off to two decimal places).

Answer 1

Correct Answer: 0.43 - 0.45

Given:
\( V(x,y,z) = 4x^2 + y^2 + z \) 

Step 1: Find the gradient
\[ \nabla V = \left( \frac{\partial V}{\partial x},\; \frac{\partial V}{\partial y},\; \frac{\partial V}{\partial z} \right) = (8x,\; 2y,\; 1) \] Step 2: Evaluate at (1,2,1)
\[ \nabla V(1,2,1) = (8,\; 4,\; 1) \] Step 3: Find magnitude
\[ |\nabla V| = \sqrt{8^2 + 4^2 + 1^2} = \sqrt{64 + 16 + 1} = \sqrt{81} = 9 \] Step 4: Unit normal vector
\[ \hat{n} = \frac{\nabla V}{|\nabla V|} = \left( \frac{8}{9},\; \frac{4}{9},\; \frac{1}{9} \right) \] Thus, \[ b = \frac{4}{9} \] Final Answer:
\[ |b| = 0.44 \]