Chapter 10: Problem 17
In Problems 15-18, compute the directional derivative of \(f(x, y)\) at the point \(P\) in the direction of the point \(Q .\) $$ f(x, y)=\sqrt{x y-2 x^{2}}, P=(1,6), Q=(3,1) $$
Short Answer
Expert verified
The directional derivative is \( \frac{-1}{2\sqrt{58}} \).
Step by step solution
01
Compute the Gradient of f(x, y)
The gradient of a function \( f(x, y) \) is a vector of its partial derivatives. First, find the partial derivative of \( f \) with respect to \( x \), \( f_x(x, y) \), and with respect to \( y \), \( f_y(x, y) \).\[ f(x, y) = \sqrt{xy - 2x^2} \]To compute \( f_x(x, y) \), use the chain rule:\[ f_x(x, y) = \frac{1}{2}\left( xy - 2x^2 \right)^{-1/2} (y - 4x) \]Similarly, compute \( f_y(x, y) \):\[ f_y(x, y) = \frac{1}{2}\left( xy - 2x^2 \right)^{-1/2} x \]Thus, the gradient is given by:\[ abla f(x, y) = \left( \frac{y - 4x}{2\sqrt{xy - 2x^2}}, \frac{x}{2\sqrt{xy - 2x^2}} \right) \]
02
Evaluate the Gradient at Point P
Substitute the coordinates of point \( P = (1, 6) \) into the gradient function \( abla f(x, y) \).\[ abla f(1, 6) = \left( \frac{6 - 4 \cdot 1}{2\sqrt{1 \cdot 6 - 2 \cdot 1^2}}, \frac{1}{2\sqrt{1 \cdot 6 - 2 \cdot 1^2}} \right) \]Calculate inside the square root first and then evaluate each component:\[ abla f(1, 6) = \left( \frac{2}{2\sqrt{2}}, \frac{1}{2\sqrt{2}} \right) = \left( \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}} \right) \]
03
Find the Direction Vector
Determine the direction vector from point \( P = (1, 6) \) to point \( Q = (3, 1) \). Subtract the coordinates of \( P \) from \( Q \):\[ \vec{d} = (3 - 1, 1 - 6) = (2, -5) \]
04
Normalize the Direction Vector
The direction for the directional derivative must be a unit vector. Calculate the magnitude of \( \vec{d} \) and divide each component by this magnitude to normalize it.\[ \| \vec{d} \| = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \]\[ \hat{d} = \left( \frac{2}{\sqrt{29}}, \frac{-5}{\sqrt{29}} \right) \]
05
Compute the Directional Derivative
The directional derivative of \( f \) at \( P \) in the direction of \( \hat{d} \) is the dot product of \( abla f(1, 6) \) and \( \hat{d} \). Compute this dot product: \[ D_{\hat{d}} f(1, 6) = abla f(1, 6) \cdot \hat{d} = \left( \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}} \right) \cdot \left( \frac{2}{\sqrt{29}}, \frac{-5}{\sqrt{29}} \right) \]Calculate each component and add them together:\[ D_{\hat{d}} f(1, 6) = \frac{2}{\sqrt{58}} + \frac{-5}{2\sqrt{58}} = \frac{4 - 5}{2\sqrt{58}} = \frac{-1}{2\sqrt{58}} \]
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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 super helpful tool when dealing with multivariable calculus. Imagine it as a magical compass pointing towards the direction where the function grows the fastest. For a function of two variables, like our example \(f(x, y)\), the gradient is a vector composed of its partial derivatives.
Here's how it looks:
The bigger the magnitude of the gradient, the steeper the path upwards.
Here's how it looks:
- Gradient vector: \( abla f(x, y) = (f_x, f_y) \)
The bigger the magnitude of the gradient, the steeper the path upwards.
Partial Derivatives
Partial derivatives are the building blocks for the gradient. They tell us how a function changes as we tweak just one variable at a time, keeping all others constant. For each variable in your equation, you can obtain a partial derivative.
In our context, the partial derivatives of \( f(x, y) = \sqrt{xy - 2x^2} \) are:
In our context, the partial derivatives of \( f(x, y) = \sqrt{xy - 2x^2} \) are:
- \(x\)-direction derivative: \( f_x(x, y) \)
- \(y\)-direction derivative: \( f_y(x, y) \)
Unit Vector
A unit vector is essentially a direction pointer, neatly normalized to have a length of one. Why use a unit vector? Because it simplifies computations while preserving direction. When calculating directional derivatives, you always use a unit vector to ensure you're getting the rate of change in the right direction.
To find a unit vector from an ordinary vector, simply divide each component by the vector's magnitude:
To find a unit vector from an ordinary vector, simply divide each component by the vector's magnitude:
- Magnitude: \( \|\vec{d}\| = \sqrt{a^2 + b^2} \)
- Normalization: \( \hat{d} = \left( \frac{a}{\|\vec{d}\|}, \frac{b}{\|\vec{d}\|} \right) \)
Dot Product
The dot product, sometimes called the scalar product, is a way of multiplying two vectors to get a number. It's like merging the two vectors into one single value, which can reveal some neat insights about their relationship.
For two-dimensional vectors, the dot product is calculated as:
For two-dimensional vectors, the dot product is calculated as:
- \( \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 \)