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Chapter 14: Problem 34

Let \(z=f(x, y), x=x(u, v), y=y(u, v)\) and \(x(1,2)=5, y(1,2)=3,\) calculate the partial derivative in terms of some of the numbers \(a, b, c, d, e, k, p, q:\) $$\begin{array}{llll} f_{x}(1,2)=a & f_{y}(1,2)=c & x_{u}(1,2)=e & y_{u}(1,2)=p \\ f_{x}(5,3)=b & f_{y}(5,3)=d & x_{v}(1,2)=k & y_{v}(1,2)=q \end{array}$$ $$z_{v}(1,2)$$

Short Answer

Expert verified
\(bk + dq\)

Step by step solution

01

Understand the Problem

We need to find the partial derivative \(z_v(1,2)\). Given the functions \(z = f(x, y), x = x(u, v), y = y(u, v)\), we can find \(z_v\) using the chain rule for multivariable calculus.
02

Apply the Chain Rule for \(z_v\)

According to the chain rule, the partial derivative \(z_v\) is computed as:\[z_v = f_x \frac{\partial x}{\partial v} + f_y \frac{\partial y}{\partial v}\]
03

Substitute Known Values

From the given information:- \(f_x(1,2) = a\), \(f_y(1,2) = c\)- New evaluated points: \(x(1,2) = 5\), \(y(1,2) = 3\) imply use \(f_x(5,3) = b\), \(f_y(5,3) = d\)- \(x_v(1,2) = k\)- \(y_v(1,2) = q\)Using this we plug into:\[z_v = f_x(5,3) \cdot x_v(1,2) + f_y(5,3) \cdot y_v(1,2)\]
04

Compute \(z_v\)

Substitute numerical values into the formula:\[z_v = b \cdot k + d \cdot q\]This results in:\[z_v = bk + dq\]
05

Conclude the Solution

The expression for \(z_v(1,2)\) in terms of the parameters is \(bk + dq\).

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Key Concepts

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

Partial Derivatives
When dealing with functions that have more than one variable, it's essential to understand partial derivatives. They help us find the rate at which the function changes with respect to one of these variables, while keeping the others constant.
This is a big change from single-variable calculus, where derivatives are straightforward. Here, you take the derivative with respect to one variable and assume the others are fixed.
  • For example, if you have a function \( f(x, y) \), the partial derivative \( f_x \) indicates how \( f \) changes as \( x \) changes and \( y \) remains constant.
  • Similarly, \( f_y \) shows the change in the function as \( y \) varies, holding \( x \) secure.

In the original exercise, we used these partial derivatives denoted by symbols \( f_x \) and \( f_y \) to compute new information about the function by combining them using the chain rule.
Multivariable Calculus
Multivariable calculus takes the principles of calculus and extends them to functions with two or more variables. This makes it quite robust, enabling analysis in higher dimensions.
It's all about analyzing functions called scalar fields, such as temperature or height, over an area.
  • A typical scenario could involve finding how a 3D shape changes its volume as one moves across it, which can be calculated using gradients, divergence, or curl operations.
  • Partial derivatives come into play again, as they are a fundamental tool in processes like optimization and curve tracing in these contexts.

The chain rule, employed in multivariable calculus, allows us to compute these derivatives with respect to multiple variable functions easily. This blending shows just how versatile calculus is beyond what simpler, single-variable types show.
Function Composition
In calculus, function composition involves selecting functions and combining them to form new functions. This is identical to the manner of crafting a story that connects different characters to enhance the plot.
In our exercise, the composition involves the function \( z = f(x, y) \), where both \( x \) and \( y \) themselves depend on \( u \) and \( v \).
  • This nested dependency is like layers of an onion, where peeling each one leads to another, crucial for seeing the complete picture.
  • To unravel these layers, we apply approaches like the chain rule, which helps find the derivative of complex compositions.

By recognizing that \( x \) and \( y \) depend on \( u \) and \( v \), we can strategically apply derivatives to each layer to compute the overall effect, as shown in the original solution. Understanding this composition is key when encountering linked systems in calculus.

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Most popular questions from this chapter

The tastiness, \(T\), of a soup depends on the volume, \(V\) of the soup in the pot and the quantity, \(S,\) of salt in the soup. If you have more soup, you need more salt to make it taste good. Match the three stories (a)-(c) to the three statements (I)-(III) about partial derivatives. (a) I started adding salt to the soup in the pot. At first the taste improved, but eventually the soup became too salty and continuing to add more salt made it worse. (b) The soup was too salty, so I started adding unsalted soup. This improved the taste at first, but eventually there was too much soup for the salt, and continuing to add unsalted soup just made it worse. (c) The soup was too salty, so adding more salt would have made it taste worse. I added a quart of unsalted soup instead. Now it is not salty enough, but I can improve the taste by adding salt. (I) \(\partial^{2} T / \partial V^{2} < 0\) (II) \(\partial^{2} T / \partial S^{2} < 0\) (III) \(\partial^{2} T / \partial V \partial S > 0\)

For \(g(x, y)\) with \(g(5,10)=100\) and \(g_{\bar{x}}(5,10)=0.5\) where \(\vec{u}\) is the unit vector in the direction of the vector \(\vec{i}+\vec{j},\) estimate \(g(5.1,10.1)\)

Find the gradient of the function. Assume the variables are restricted to a domain on which the function s defined. $$z=(x+y) e^{y}$$

Find the gradient of the function. Assume the variables are restricted to a domain on which the function s defined. $$z=\sin (x / y)$$

(a) Find the differential of \(g(u, v)=u^{2}+u v\) (b) Use your answer to part (a) to estimate the change in \(g\) as you move from (1,2) to (1.2,2.1)
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