The directional derivative of the function 𝑓 given below at the point (1,0) in the direction of 1/2 (𝑖+√3 𝑗) is ____________ (rounded off to 1 decimal place).

𝑓(𝑥,𝑦)= x² + 𝑥𝑦²

The correct answer is 1.

Step-by-step Explanation:

To find the directional derivative of the function 𝑓(𝑥,𝑦) at a given point in a specific direction, we use the gradient of the function and the unit direction vector.

1. Identify the given function and point:
The function is:
$f(x,y)=x^2+x{y}^2$
The given point is (1,0).

2. Find the gradient of the function, ∇����(𝑥,𝑦):
The gradient is a vector consisting of the first-order partial derivatives of the function with respect to 𝑥 and 𝑦:
$\nabla f(x,y)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$
Let us compute the partial derivatives:
• Partial derivative with respect to 𝑥:
$\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}(x^2+x{y}^2)=2x+y^2$
• Partial derivative with respect to 𝑦:
$\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}(x^2+x{y}^2)=2xy$

3. Evaluate the gradient at the point (1,0):
Substitute 𝑥 = 1 and 𝑦 = 0 into the partial derivatives:
$\frac{\partial f}{\partial x}{|}_{(1,0)}=2\left(1\right)+0^2=2$
$\frac{\partial f}{\partial y}{|}_{(1,0)}=2\left(1\right)\left(0\right)=0$
Thus, the gradient vector at (1,0) is:
$\nabla f(1,0)=(2,0)=2i+0j$

4. Check the direction vector:
The given direction vector is:
$u=\frac{1}{2}(i+\sqrt{3}j)=\frac{1}{2}i+\frac{\sqrt{3}}{2}j$
Let us verify if it is a unit vector by calculating its magnitude:
$\parallel u\parallel =\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=\sqrt{1}=1$
Since the magnitude is 1, it is already a unit vector.

5. Compute the directional derivative:
The directional derivative of 𝑓 at the point (1,0) in the direction of the unit vector 𝑢 is the dot product of the gradient and the unit vector:
$D_{u}f(1,0)=\nabla f(1,0)·u$
$D_{u}f(1,0)=(2i+0j)·(\frac{1}{2}i+\frac{\sqrt{3}}{2}j)$
$D_{u}f(1,0)=2·\frac{1}{2}+0·\frac{\sqrt{3}}{2}=1$

Thus, the directional derivative is exactly 1 (or 1.0 rounded to one decimal place).