Chapter 14: Problem 31
Zero directional derivative In what direction is the derivative of \(f(x, y)=x y+y^{2}\) at \(P(3,2)\) equal to zero?
Short Answer
Expert verified
The derivative is zero in the directions \((-\frac{7}{\sqrt{53}},\frac{2}{\sqrt{53}})\) and \((\frac{7}{\sqrt{53}}, -\frac{2}{\sqrt{53}})\).
Step by step solution
01
Compute the Gradient of the Function
The gradient of a function \(f(x, y)\) is given by \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). First, compute \(\frac{\partial f}{\partial x}\):\[\frac{\partial f}{\partial x} = y\].Next, compute \(\frac{\partial f}{\partial y}\): \[\frac{\partial f}{\partial y} = x + 2y\].Thus, the gradient is \(abla f = (y, x + 2y)\).
02
Evaluate the Gradient at Point P
Substitute \(P(3,2)\) into the gradient \((y, x + 2y)\) to find \(abla f(3,2)\):- For \(y\): \(y = 2\).- For \(x + 2y\): \(3 + 2 \times 2 = 3 + 4 = 7\).The gradient at \(P(3,2)\) is \((2, 7)\).
03
Determine the Directional Derivative
The directional derivative of a function \(f\) at a point \(P\) in the direction of a unit vector \(\mathbf{u} = (u_1, u_2)\) is given by \(D_{\mathbf{u}}f = abla f \cdot \mathbf{u}\).This means:\[D_{\mathbf{u}}f = 2u_1 + 7u_2\].
04
Set Directional Derivative to Zero and Solve
To ensure the directional derivative is zero, set \(2u_1 + 7u_2 = 0\). Solve this equation for \((u_1, u_2)\):\[2u_1 = -7u_2\]Express \(u_1\) in terms of \(u_2\):\[u_1 = -\frac{7}{2}u_2\].
05
Find Unit Vector
Since \((u_1, u_2)\) should be a unit vector, it must satisfy \(u_1^2 + u_2^2 = 1\).Substitute \(u_1 = -\frac{7}{2}u_2\) into the equation:\[\left( -\frac{7}{2}u_2 \right)^2 + u_2^2 = 1\],\[\frac{49}{4}u_2^2 + u_2^2 = 1\],\[\frac{53}{4}u_2^2 = 1\],\[u_2^2 = \frac{4}{53}\],\[u_2 = \pm\frac{2}{\sqrt{53}}\].
06
Determine the Corresponding u1
Use \(u_1 = -\frac{7}{2}u_2\) to find \(u_1\):- If \(u_2 = \frac{2}{\sqrt{53}}\), then \(u_1 = -\frac{7}{2}\times\frac{2}{\sqrt{53}} = -\frac{7}{\sqrt{53}}\).- If \(u_2 = -\frac{2}{\sqrt{53}}\), then \(u_1 = -\frac{7}{2}\times\left(-\frac{2}{\sqrt{53}}\right) = \frac{7}{\sqrt{53}}\).
07
Conclude the Directions
The unit vectors \(\left(-\frac{7}{\sqrt{53}}, \frac{2}{\sqrt{53}}\right)\) and \(\left(\frac{7}{\sqrt{53}}, -\frac{2}{\sqrt{53}}\right)\) both yield a directional derivative of zero at point \(P(3,2)\). These are the directions where the derivative of \(f\) is zero.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
The gradient of a function is a crucial concept when exploring directional derivatives. It represents how a function changes at any given point and directs us toward the steepest ascent. In two dimensions, for a function of two variables such as \( f(x, y) \), the gradient \( abla f \) is a vector consisting of the function's partial derivatives with respect to each variable.
The gradient is noted as:
The gradient is noted as:
- \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)
- \( \frac{\partial f}{\partial x} = y \)
- \( \frac{\partial f}{\partial y} = x + 2y \)
Partial Derivatives
Partial derivatives are foundational when working with functions of multiple variables. They allow us to investigate how a function changes with respect to one variable while keeping others constant. In essence, partial derivatives are just derivatives in multi-variable calculus, focusing on one dimension of the problem at a time.
For the function \( f(x, y) = xy + y^2 \), its partial derivatives are:
For the function \( f(x, y) = xy + y^2 \), its partial derivatives are:
- With respect to \( x \): \( \frac{\partial f}{\partial x} = y \)
- With respect to \( y \): \( \frac{\partial f}{\partial y} = x + 2y \)
Unit Vectors
Understanding unit vectors is integral to calculating directional derivatives. A unit vector has a magnitude of 1 and is used to indicate direction. It plays a vital role in ensuring that when computing directional derivatives, we are analyzing shifts that are solely due to direction, not length.
A general unit vector \( \mathbf{u} \) in two dimensions is represented as \( (u_1, u_2) \), meeting the condition:
A general unit vector \( \mathbf{u} \) in two dimensions is represented as \( (u_1, u_2) \), meeting the condition:
- \( u_1^2 + u_2^2 = 1 \)