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Magazine Chapter 13: Problem 26
Let $$ f(x, y)=\frac{y}{x+y} $$ Find a unit vector u for which \(D_{\mathbf{u}} f(2,3)=0\)
Short Answer
Expert verified
The unit vector \( \mathbf{u} = \langle \frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \rangle \) or \( \mathbf{u} = \langle -\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}} \rangle \).
Step by step solution
01
Calculate Partial Derivatives
First, we need to find the partial derivatives of the function \( f(x, y) = \frac{y}{x+y} \). The partial derivative with respect to \( x \) is given by \[ f_x = \frac{\partial}{\partial x} \left( \frac{y}{x+y} \right) = \frac{-y}{(x+y)^2} \]The partial derivative with respect to \( y \) is given by \[ f_y = \frac{\partial}{\partial y} \left( \frac{y}{x+y} \right) = \frac{x}{(x+y)^2} \]
02
Evaluate the Partial Derivatives at the Given Point
Now, we evaluate the partial derivatives at the point \( (2, 3) \):\[ f_x(2, 3) = \frac{-3}{(2+3)^2} = \frac{-3}{25} \]\[ f_y(2, 3) = \frac{2}{(2+3)^2} = \frac{2}{25} \]
03
Set Directional Derivative to Zero
To find the unit vector \( \mathbf{u} = \langle a, b \rangle \) for which \( D_{\mathbf{u}} f(2, 3) = 0 \), use the directional derivative formula:\[ D_{\mathbf{u}} f(2, 3) = f_x(2, 3) \cdot a + f_y(2, 3) \cdot b \]Set this equal to zero:\[ \frac{-3}{25} \cdot a + \frac{2}{25} \cdot b = 0 \]
04
Solve for the Ratio of a to b
To solve \( \frac{-3}{25} a + \frac{2}{25} b = 0 \), simplify it:\[ -3a + 2b = 0 \]Thus, \( 2b = 3a \), leading to:\[ \frac{b}{a} = \frac{3}{2} \]
05
Find the Unit Vector
Since the vector \( \langle a, b \rangle \) must be a unit vector, \[ \sqrt{a^2 + b^2} = 1 \]Using \( b = \frac{3}{2}a \) in the unit vector equation:\[ \sqrt{a^2 + \left( \frac{3}{2}a \right)^2} = 1 \]Simplify to find \( a \):\[ \sqrt{a^2 + \frac{9}{4}a^2} = 1 \]\[ \sqrt{\frac{13}{4}a^2} = 1 \]\[ \frac{\sqrt{13}}{2}|a| = 1 \]\[ |a| = \frac{2}{\sqrt{13}} \]Thus, \( a = \frac{2}{\sqrt{13}} \) or \( a = -\frac{2}{\sqrt{13}} \). Now, find \( b \):If \( a = \frac{2}{\sqrt{13}} \), then:\[ b = \frac{3}{2} \cdot \frac{2}{\sqrt{13}} = \frac{3}{\sqrt{13}} \]The unit vector can be \( \mathbf{u} = \langle \frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \rangle \) or its reflection \( \mathbf{u} = \langle -\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}} \rangle \).
06
Verify the Unit Vector
To verify, check:\[ \left( \frac{2}{\sqrt{13}} \right)^2 + \left( \frac{3}{\sqrt{13}} \right)^2 = \frac{4}{13} + \frac{9}{13} = 1 \]So, \( \mathbf{u} = \langle \frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}} \rangle \) is indeed a unit vector. Similarly,\[ \left( -\frac{2}{\sqrt{13}} \right)^2 + \left( -\frac{3}{\sqrt{13}} \right)^2 = \frac{4}{13} + \frac{9}{13} = 1 \]This confirms both vectors are valid unit vectors.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives help us understand how a function changes when we vary one variable, keeping the other constant. For a function of two variables, like our case with \( f(x, y) = \frac{y}{x+y} \), we can calculate two partial derivatives: with respect to \(x\) and \(y\).
This provides critical information when determining directional derivatives, which measure the rate of change of the function in any direction.
- The partial derivative with respect to \(x\), noted as \( f_x \), treats \( y \) as a constant. It examines how small changes in \( x \) affect the function.
- The partial derivative with respect to \(y\), \( f_y \), keeps \( x \) constant and looks at how variations in \( y \) impact \( f(x, y) \).
This provides critical information when determining directional derivatives, which measure the rate of change of the function in any direction.
Unit Vector
A unit vector has a length of one and specifies a direction in a plane or space. When dealing with directional derivatives, converting a direction into a unit vector ensures the derivative measures the rate of change per unit length.
To find a unit vector, \( \mathbf{u} = \langle a, b \rangle \), the condition \( \sqrt{a^2 + b^2} = 1 \) must hold true.
To find a unit vector, \( \mathbf{u} = \langle a, b \rangle \), the condition \( \sqrt{a^2 + b^2} = 1 \) must hold true.
- For our exercise, we find that the required unit vector also needs to make the directional derivative zero. We achieve this by solving \( \frac{-3}{25} \cdot a + \frac{2}{25} \cdot b = 0 \).
- Once the relationship between \( a \) and \( b \) is established, ensuring the vector is a unit vector demands solving \( \sqrt{a^2 + b^2} = 1 \) using the ratio \( \frac{b}{a} = \frac{3}{2} \).
Gradient
The gradient is a vector that points in the direction of the maximum rate of increase of a function, composed of the partial derivatives. For a function \( f(x, y) \), the gradient is \( abla f = \langle f_x, f_y \rangle \). It serves as an essential tool in understanding directional derivatives.
- In our exercise, the gradient provides the weights for how \( a \) and \( b \) in the unit vector influence the directional derivative.
- In essence, the directional derivative is a dot product of the gradient and the unit vector: \( D_{\mathbf{u}} f(x, y) = abla f \cdot \mathbf{u} \).
