Question

Let \(\vec{v} = 4\hat{i} + 3\hat{k}\) be a vector, where \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) represent unit vectors along the axes \(x\), \(y\) and \(z\) respectively. Then, the directional derivative of the scalar function \(f(x,y,z) = 2 \ln(xy) + \ln(yz) + 3\ln(zx)\) at the point \((1,1,1)\) in the direction of \(\vec{v}\) is _______

• A. 7.2
• B. 6
• C. 6.2
• D. 6.4

Hint:

Always simplify logarithmic functions before taking partial derivatives. It prevents chain rule complexity and dramatically reduces computation errors.
Don't forget to normalize the direction vector \(\vec{v}\) to a unit vector \(\hat{u}\) before calculating the dot product.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks us to calculate the directional derivative of a multi-variable scalar function \(f(x,y,z)\) at a specific point \((1,1,1)\) in the direction of a given vector \(\vec{v}\).
The directional derivative measures the rate of change of the function per unit distance along that specific direction.

Step 2: Key Formula or Approach:
The directional derivative \(D_{\hat{u}} f\) of a scalar function \(f\) in the direction of a unit vector \(\hat{u}\) is calculated using the gradient of the function:
\[ D_{\hat{u}} f = \nabla f \cdot \hat{u} \] where:
\(\nabla f = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k}\) is the gradient vector of \(f\).
\(\hat{u} = \frac{\vec{v}}{|\vec{v}|}\) is the unit vector in the direction of \(\vec{v}\).

Step 3: Detailed Explanation:
Let us simplify the scalar function \(f(x,y,z)\) first using the properties of logarithms to make differentiation easier:
\[ f(x,y,z) = 2 \ln(xy) + \ln(yz) + 3\ln(zx) \] Using \(\ln(ab) = \ln(a) + \ln(b)\):
\[ f(x,y,z) = 2(\ln x + \ln y) + (\ln y + \ln z) + 3(\ln z + \ln x) \] Group the terms for each variable:
\[ f(x,y,z) = (2\ln x + 3\ln x) + (2\ln y + \ln y) + (\ln z + 3\ln z) \] \[ f(x,y,z) = 5\ln x + 3\ln y + 4\ln z \] Now, compute the partial derivatives of \(f\):
\[ \frac{\partial f}{\partial x} = \frac{5}{x} \] \[ \frac{\partial f}{\partial y} = \frac{3}{y} \] \[ \frac{\partial f}{\partial z} = \frac{4}{z} \] Evaluate these partial derivatives at the given point \((1,1,1)\):
\[ \frac{\partial f}{\partial x} \Big|_{(1,1,1)} = 5 \] \[ \frac{\partial f}{\partial y} \Big|_{(1,1,1)} = 3 \] \[ \frac{\partial f}{\partial z} \Big|_{(1,1,1)} = 4 \] Construct the gradient vector \(\nabla f\) at \((1,1,1)\):
\[ \nabla f(1,1,1) = 5\hat{i} + 3\hat{j} + 4\hat{k} \] Next, find the unit vector \(\hat{u}\) in the direction of \(\vec{v} = 4\hat{i} + 3\hat{k}\):
The magnitude of \(\vec{v}\) is:
\[ |\vec{v}| = \sqrt{4^2 + 0^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] The unit direction vector is:
\[ \hat{u} = \frac{4\hat{i} + 3\hat{k}}{5} \] Finally, calculate the directional derivative by taking the dot product of \(\nabla f(1,1,1)\) and \(\hat{u}\):
\[ D_{\hat{u}} f = (5\hat{i} + 3\hat{j} + 4\hat{k}) \cdot \left(\frac{4\hat{i} + 3\hat{k}}{5}\right) \] \[ D_{\hat{u}} f = \frac{(5 \times 4) + (3 \times 0) + (4 \times 3)}{5} \] \[ D_{\hat{u}} f = \frac{20 + 0 + 12}{5} = \frac{32}{5} = 6.4 \] This matches Option (D) perfectly.

Step 4: Final Answer
Thus, the directional derivative is \(6.4\), which corresponds to option (D).