Question

Find two unit vectors u such that Dᵤf(1,1)=0. Here, f(x,y)=x²+9y².

Answer 1

Two unit vectors
`\( \textbf{u}_1 \) and \( \textbf{u}_2 \)` that satisfy
`\( D_{\textbf{u}_1}f(1,1) = 0 \) and \( D_{\textbf{u}_2}f(1,1) = 0 \) are \( \textbf{u}_1 = (1)/(√(2))\langle -√(2), 1 \rangle \) and \( \textbf{u}_2 = (1)/(√(2))\langle √(2), 1 \rangle \).`

Step-by-step explanation:

The directional derivative of a function
`\( f(x, y) \)` at a point
`\( (x_0, y_0) \)` in the direction of a unit vector
`\( \textbf{u} = \langle a, b \rangle \)` is given by
`\( D_{\textbf{u}}f(x_0, y_0) = (\partial f)/(\partial x)a + (\partial f)/(\partial y)b \).`

For the function
`\( f(x, y) = x^2 + 9y^2 \),` the partial derivatives are
`\( (\partial f)/(\partial x) = 2x \) and \( (\partial f)/(\partial y) = 18y \)`. At the point ( (1, 1) ), these partial derivatives evaluate to
`\( (\partial f)/(\partial x)(1, 1) = 2 \) and \( (\partial f)/(\partial y)(1, 1) = 18 \).`

To find unit vectors
`\( \textbf{u} = \langle a, b \rangle \) such that \( D_{\textbf{u}}f(1,1) = 0 \)`, we set up the equation
`\( (\partial f)/(\partial x)(1, 1) \cdot a + (\partial f)/(\partial y)(1, 1) \cdot b = 0 \)` and solve for ( a ) and ( b ) under the condition that
`\( a^2 + b^2 = 1 \) (as \( \textbf{u} \) is a unit vector).`

Solving this system of equations yields two unit vectors:

`\( \textbf{u}_1 = (1)/(√(2))\langle -√(2), 1 \rangle \) and \( \textbf{u}_2 = (1)/(√(2))\langle √(2), 1 \rangle \), both of which satisfy \( D_{\textbf{u}_1}f(1,1) = 0 \) and \( D_{\textbf{u}_2}f(1,1) = 0 \).`