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Chapter 15: Problem 27

Computing directional derivatives with the gradient Compute the directional derivative of the following functions at the given point \(P\) in the direction of the given vector. Be sure to use a unit vector for the direction vector. $$f(x, y)=3 x^{2}+2 y+5 ; P(1,2) ;\langle-3,4\rangle$$

Short Answer

Expert verified
Answer: The directional derivative is \(-2\).

Step by step solution

01

Compute the unit vector of direction vector

To find the unit vector of the given direction vector, we first calculate its length. For \(\langle -3, 4 \rangle\), the length is: $$ \parallel \langle -3, 4 \rangle \parallel = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$ Therefore, the unit vector is: $$ \frac{\langle -3, 4 \rangle}{\parallel \langle -3, 4 \rangle \parallel} = \frac{\langle -3, 4 \rangle}{5} = \langle -\frac{3}{5}, \frac{4}{5} \rangle $$
02

Find the gradient of the function

To find the gradient of the function \(f(x, y) = 3x^2 + 2y + 5\), we compute the partial derivatives with respect to \(x\) and \(y\): $$ \frac{\partial f}{\partial x} = 6x \quad \text{and} \quad \frac{\partial f}{\partial y} = 2 $$ Now, we can find the gradient of the function at point \(P(1, 2)\): $$ \nabla f(1, 2) = \langle \frac{\partial f}{\partial x}(1, 2), \frac{\partial f}{\partial y}(1, 2) \rangle = \langle 6(1), 2 \rangle = \langle 6, 2 \rangle $$
03

Compute the directional derivative

Now, we can compute the directional derivative of the function \(f(x, y) = 3x^2 + 2y + 5\) at point \(P(1, 2)\) in the direction of the given unit vector \(\langle -\frac{3}{5}, \frac{4}{5} \rangle\). The directional derivative can be found by taking the dot product between the gradient of the function at point \(P\) and the unit vector: $$ D_{\langle -\frac{3}{5}, \frac{4}{5} \rangle} f(1, 2) = \nabla f(1, 2) \cdot \langle -\frac{3}{5}, \frac{4}{5} \rangle = \langle 6, 2 \rangle \cdot \langle -\frac{3}{5}, \frac{4}{5} \rangle $$ $$ D_{\langle -\frac{3}{5}, \frac{4}{5} \rangle} f(1, 2) = 6 \cdot (-\frac{3}{5}) + 2 \cdot \frac{4}{5} = -\frac{18}{5} + \frac{8}{5} = -\frac{10}{5} $$ The directional derivative of the function \(f(x, y) = 3x^2 + 2y + 5\) at point \(P(1, 2)\) in the direction of the vector \(\langle -3, 4 \rangle\) is: $$ D_{\langle -\frac{3}{5}, \frac{4}{5} \rangle} f(1, 2) = -\frac{10}{5} = -2 $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gradient Vector
Understanding the gradient vector is essential for computing directional derivatives. In multivariable calculus, the gradient is a vector that contains all the partial derivatives of a function. It points in the direction of the greatest rate of increase of the function, and its magnitude denotes the rate of increase in that direction.

For a function of two variables, like the function in our exercise, which is expressed as \( f(x, y) = 3x^2 + 2y + 5 \), the gradient vector would include partial derivatives with respect to each variable. Once we compute \( \frac{\partial f}{\partial x} = 6x \) and \( \frac{\partial f}{\partial y} = 2 \) and evaluate them at a specific point鈥攈ere, \( P(1, 2) \)鈥攚e get the gradient vector \( abla f(1, 2) = \langle 6, 2 \rangle \).
Unit Vector
A unit vector is a vector with a length, or magnitude, of 1. It indicates a direction in space, and in the context of directional derivatives, it specifies the direction in which the rate of change is measured. To find a unit vector, we divide a given vector by its magnitude.

In the solution provided, the unit vector of \( \langle -3, 4 \rangle \) is calculated by dividing the vector by its length: \( \frac{\langle -3, 4 \rangle}{5} = \langle -\frac{3}{5}, \frac{4}{5} \rangle \). This unit vector is crucial because it standardizes the directional derivative, ensuring the measure of the rate of change is independent of the length of the original direction vector.
Partial Derivatives
Partial derivatives represent how a function changes as one variable changes while all other variables are held constant. They are fundamental to the computation of the gradient vector and directional derivatives in multivariable functions.

For our function \( f(x, y) = 3x^2 + 2y + 5 \), we compute partial derivatives with respect to \( x \) and \( y \) to find how the function's output changes along each axis. The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 6x \), indicating that for any change in \( x \), the output changes by 6 times the change in \( x \). Similarly, the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2 \), indicating a constant change along the \( y \)-axis, regardless of \( y \)'s value.

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Most popular questions from this chapter

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