/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Define the derivative of the fun... [FREE SOLUTION] | 91Ó°ÊÓ

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Define the derivative of the function \(z=f(x, y)\) in the direction \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\)

Short Answer

Expert verified
The directional derivative of \(f(x, y)\) in the direction of \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\) is given by the formula \(D_{\mathbf{u}}f(x, y) = f_x(x, y)\cos \theta + f_y(x, y)\sin \theta\). However, without a specified function \(f(x, y)\), we can't calculate a specific expression.

Step by step solution

01

Understanding the definition of a directional derivative

The directional derivative of a function \(z=f(x, y)\) in the direction of a unit vector \(\mathbf{u}\) can be found using the equation \(D_{\mathbf{u}}f(x, y) = f_x(x, y)\cos \theta + f_y(x, y)\sin \theta\), where \(f_x(x, y)\) and \(f_y(x, y)\) are the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\), respectively, and \(\theta\) is the angle the direction vector \(\mathbf{u}\) makes with the positive x-axis.
02

Calculate partial derivatives

The exercise only defines the form of \(f(x, y)\) generally without specifying the function, so it's impossible to give a specific solution for \(f_x(x, y)\) and \(f_y(x, y)\). It could be executed if \(f(x, y)\) was given.
03

Plug the partial derivatives and direction into the formula

Again, without specific functions for \(f_x(x, y)\) and \(f_y(x, y)\), we can't evaluate this step completely. However, assuming that we could compute those, we would then substitute those values into our formula for the directional derivative, along with the direction \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\). The result would be an expression involving \(\cos(\theta)\) and \(\sin(\theta)\), whose specific form would depend on the original function \(f(x, y)\).

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Key Concepts

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

Partial Derivatives
When we talk about partial derivatives in the magic world of multivariable calculus, we're dealing with functions that have more than one input variable. Imagine we have a function, like a recipe, that takes multiple ingredients ( values) and gives us a delicious dish (output). But sometimes, you want to know how changing one ingredient a little bit, while keeping all the others fixed, affects the final taste of the dish. That's exactly what partial derivatives do for us in math.

They tell us how sensitive our function (like the taste of our dish) is to tiny changes in just one of the input variables (imagine just a pinch more salt). When we write down text, the small letter with the subscript looks like this: ( ). We're picking one variable at a time and asking how the function reacts to changes along that axis alone.

For instance, in our exercise considering a function ( ), let's say the 'x' ingredient. If we want to find out how this function changes with a slight tweak in 'x', we calculate the partial derivative with respect to 'x', denoted as ( ). Similarly, to check the impact of changing 'y', we look at ( ). Once we get these, we're armed with the knowledge of how sensitive our function is to each of the ingredients—be it 'x' or 'y'. This is like understanding the individual roles of salt and pepper in your dish.
Multivariable Calculus
If partial derivatives are about adjusting one ingredient at a time, then multivariable calculus is like the whole cooking show, where we deal with all the ingredients at once. Imagine you're not just interested in what happens if you tweak salt or pepper individually, but also when you season your dish with both at the same time and then also add some herbs.

This branch of mathematics allows us to study functions that depend on several variables. It's like a dance of many elements, and it gives us tools, like the directional derivative from our exercise, to see exactly how our function will change if we move in a very specific direction, like adding a combination of spices to a dish in a precise manner.

Moving beyond mere partial derivatives, multivariable calculus involves gradient vectors (which point in the direction of the steepest ascent), optimization of functions with many variables (such as finding the perfect balance of ingredients for the most flavor), and even multiple integrals (like adding up all the tiny flavor contributions from each ingredient to get the overall taste). These are just a few of the delectable concepts in the smorgasbord that is multivariable calculus. It enriches our understanding much like how complex flavors enrich a dish.
Vector Calculus
Vector calculus is the cherry on top in our delicious math sundae. It deals not only with quantities but also with their direction. What's a direction in math? Well, just like a recipe might instruct you to 'stir clockwise', in vector calculus, 'direction' tells us more than just 'how much'. It tells us 'how' and 'where' to proceed to achieve the desired outcome.

When we dive into vector calculus, we meet things like vector fields, which are like assigning a specific stirring action to every point in our batter (or every point in space, to be exact). How does that apply to our exercise? The directional derivative is a fancy term for knowing exactly what happens to our function ( ) when we nudge it a little in a very specific direction—not just any direction. The vector makes sure we're stirring our function exactly the way we want, preserving the essence of the flavors, i.e., the function's properties, in that direction.

...n � is a bit like whisking that batter in just the right direction: too much and you're overworking the mixture, too little and it's lump city.

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Most popular questions from this chapter

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