Question

Consider the function \( z = x^3 - 2x^2y + xy^2 + 1 \). The directional derivative of z at the point (1, 2) along the direction \( 3\hat{i} + 4\hat{j} \) is

• A. 0
• B. -1
• C. 1
• D. -2

Hint:

A common mistake is to forget to normalize the direction vector. The directional derivative is defined with respect to a unit vector, so always divide the given direction vector by its magnitude before taking the dot product.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The directional derivative of a function \(z(x, y)\) at a point \((x_0, y_0)\) in the direction of a unit vector \( \vec{u} \) measures the rate of change of the function at that point as one moves in the direction of \( \vec{u} \). It is calculated as the dot product of the gradient of the function at that point and the unit direction vector.
Step 2: Key Formula or Approach:
The directional derivative \(D_{\vec{u}}z\) is given by:
\[ D_{\vec{u}}z = (\nabla z) . \vec{u} \] 1. Calculate the gradient of z, \( \nabla z = \frac{\partial z}{\partial x}\hat{i} + \frac{\partial z}{\partial y}\hat{j} \).
2. Evaluate the gradient at the given point (1, 2).
3. Find the unit vector \( \vec{u} \) in the given direction.
4. Compute the dot product.
Step 3: Detailed Calculation:
1. Calculate the gradient \( \nabla z \):
The function is \( z = x^3 - 2x^2y + xy^2 + 1 \).
\[ \frac{\partial z}{\partial x} = 3x^2 - 4xy + y^2 \] \[ \frac{\partial z}{\partial y} = -2x^2 + 2xy \] So, \( \nabla z = (3x^2 - 4xy + y^2)\hat{i} + (-2x^2 + 2xy)\hat{j} \).
2. Evaluate the gradient at (1, 2):
Substitute \(x=1\) and \(y=2\).
\[ \frac{\partial z}{\partial x}\bigg|_{(1,2)} = 3(1)^2 - 4(1)(2) + (2)^2 = 3 - 8 + 4 = -1 \] \[ \frac{\partial z}{\partial y}\bigg|_{(1,2)} = -2(1)^2 + 2(1)(2) = -2 + 4 = 2 \] So, \( \nabla z \bigg|_{(1,2)} = -1\hat{i} + 2\hat{j} \).
3. Find the unit vector \( \vec{u} \):
The given direction vector is \( \vec{v} = 3\hat{i} + 4\hat{j} \).
The magnitude of \( \vec{v} \) is \( |\vec{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).
The unit vector is \( \vec{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{3\hat{i} + 4\hat{j}}{5} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} \).
4. Compute the dot product:
\[ D_{\vec{u}}z = (\nabla z) . \vec{u} = (-1\hat{i} + 2\hat{j}) . \left(\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}\right) \] \[ D_{\vec{u}}z = (-1)\left(\frac{3}{5}\right) + (2)\left(\frac{4}{5}\right) = -\frac{3}{5} + \frac{8}{5} = \frac{5}{5} = 1 \] Step 4: Final Answer:
The directional derivative is 1.