Problem 17 - Calculus for Biology and Medicine

Question

In Problems 15-18, compute the directional derivative of \(f(x, y)\) at the point \(P\) in the direction of the point \(Q .\) $$ f(x, y)=\sqrt{x y-2 x^{2}}, P=(1,6), Q=(3,1) $$

Step-by-Step Solution

Answer

The directional derivative is \( \frac{-1}{2\sqrt{58}} \).

Step 1: 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:\[ abla f(x, y) = \left( \frac{y - 4x}{2\sqrt{xy - 2x^2}}, \frac{x}{2\sqrt{xy - 2x^2}} \right) \]

Step 2: 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:\[ abla f(1, 6) = \left( \frac{2}{2\sqrt{2}}, \frac{1}{2\sqrt{2}} \right) = \left( \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}} \right) \]

Step 3: Find the Direction Vector

Determine the direction vector from point \( P = (1, 6) \) to point \( Q = (3, 1) \). Subtract the coordinates of \( P \) from \( Q \):\[ \vec{d} = (3 - 1, 1 - 6) = (2, -5) \]

Step 4: 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.\[ \| \vec{d} \| = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \]\[ \hat{d} = \left( \frac{2}{\sqrt{29}}, \frac{-5}{\sqrt{29}} \right) \]

Step 5: 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:\[ D_{\hat{d}} f(1, 6) = \frac{2}{\sqrt{58}} + \frac{-5}{2\sqrt{58}} = \frac{4 - 5}{2\sqrt{58}} = \frac{-1}{2\sqrt{58}} \]

Key Concepts

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:

• Gradient vector: \( abla f(x, y) = (f_x, f_y) \)

Both components, \( f_x \) and \( f_y \), give us information on how \( f \) changes as we move along the \( x \) and \( y \) axes, respectively. This vector not only tells us the direction of the biggest increase but also gives an idea about the rate of change in that direction.
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:

• \(x\)-direction derivative: \( f_x(x, y) \)
• \(y\)-direction derivative: \( f_y(x, y) \)

Think of these like tuning knobs: twisting each one tells you how the function behaves if you adjust just one input, making them indispensable for understanding the dynamics of multivariable relationships.

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:

• Magnitude: \( \|\vec{d}\| = \sqrt{a^2 + b^2} \)
• Normalization: \( \hat{d} = \left( \frac{a}{\|\vec{d}\|}, \frac{b}{\|\vec{d}\|} \right) \)

This ensures that no matter the size of your original vector, your unit vector will always have a length of one.

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:

• \( \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 \)

In directional derivatives, we use the dot product of the gradient and the unit vector. What this does is project the gradient onto the direction of interest, giving a scalar that tells us the rate at which the function increases in that exact direction. It's like checking if you're walking briskly uphill or ambling gently.