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\(39-42\) Use implicit differentiation to find \(\partial z / \partial x\) and \(\partial z / \partial y\) $$x^{2}+2 y^{2}+3 z^{2}=1$$

Short Answer

Expert verified
\( \frac{\partial z}{\partial x} = -\frac{x}{3z} \) and \( \frac{\partial z}{\partial y} = -\frac{2y}{3z} \).

Step by step solution

01

Differentiate with respect to x implicitly

Differentiate the entire equation \( x^2 + 2y^2 + 3z^2 = 1 \) with respect to \( x \). Remember to use the chain rule for the \( z^2 \) term.\( \frac{d}{dx}(x^2) + \frac{d}{dx}(2y^2) + \frac{d}{dx}(3z^2) = \frac{d}{dx}(1) \)
02

Compute the derivatives

Calculate the derivatives for each term from Step 1.\( 2x + 0 + 3(2z)\frac{\partial z}{\partial x} = 0 \). The \( y^2 \) term vanishes since \( y \) does not depend on \( x \). Simplify this to \( 2x + 6z\frac{\partial z}{\partial x} = 0 \).
03

Solve for \( \frac{\partial z}{\partial x} \)

Rearrange the equation from Step 2 to solve for \( \frac{\partial z}{\partial x} \).\[ 6z \frac{\partial z}{\partial x} = -2x \]. Divide both sides by \( 6z \) to isolate \( \frac{\partial z}{\partial x} \).\( \frac{\partial z}{\partial x} = -\frac{x}{3z} \).
04

Differentiate with respect to y implicitly

Differentiate the entire equation \( x^2 + 2y^2 + 3z^2 = 1 \) with respect to \( y \). Use the chain rule for the \( z^2 \) term.\( \frac{d}{dy}(x^2) + \frac{d}{dy}(2y^2) + \frac{d}{dy}(3z^2) = \frac{d}{dy}(1) \)
05

Compute the derivatives

Calculate the derivatives for each term from Step 4.\( 0 + 4y + 3(2z)\frac{\partial z}{\partial y} = 0 \). The \( x^2 \) term vanishes since \( x \) does not depend on \( y \). Simplify this to \( 4y + 6z\frac{\partial z}{\partial y} = 0 \).
06

Solve for \( \frac{\partial z}{\partial y} \)

Rearrange the equation from Step 5 to solve for \( \frac{\partial z}{\partial y} \).\[ 6z \frac{\partial z}{\partial y} = -4y \]. Divide both sides by \( 6z \) to isolate \( \frac{\partial z}{\partial y} \).\( \frac{\partial z}{\partial y} = -\frac{2y}{3z} \).

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Key Concepts

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

Partial Derivatives
Partial derivatives allow us to understand how a multivariable function changes with respect to one of its variables, while keeping the others constant. For a function like the equation given, \(x^2 + 2y^2 + 3z^2 = 1\), the goal is to find how \(z\) changes in response to changes in \(x\) and \(y\). Here, we are interested in \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).
By implicitly differentiating the equation:
  • We treat \(z\) as a function of \(x\) and \(y\), allowing us to compute how changes in either \(x\) or \(y\) affect \(z\).
  • The partial derivative \(\frac{\partial z}{\partial x}\) means we differentiate concerning \(x\) while treating \(y\) as a constant. Similarly, \(\frac{\partial z}{\partial y}\) treats \(x\) as a constant while differentiating concerning \(y\).
These derivatives are vital in many fields such as economics and physics, where outcomes often depend on multiple changing factors.
Chain Rule
The chain rule in calculus is an essential tool for differentiating compositions of functions. When finding partial derivatives in multivariable functions using implicit differentiation, the chain rule helps in managing the dependencies between variables.
Here, we encounter the chain rule while differentiating terms involving \(z^2\). Since \(z\) depends on \(x\) and \(y\), the differentiation of \(3z^2\) requires applying the chain rule:
  • We differentiate \(3z^2\) with respect to \(x\) as \(6z\frac{\partial z}{\partial x}\).
  • Similarly, differentiating \(3z^2\) with respect to \(y\) gives \(6z\frac{\partial z}{\partial y}\).
These applications are vital because they allow us to incorporate the way one variable influences another into our derivative calculations. Often in multivariable contexts, variables do not change independently, making the chain rule indispensable for implicit differentiation.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of multiple variables. It explores how these functions change when one or more of the variables change. In the context of the given function \(x^2 + 2y^2 + 3z^2 = 1\), we work with three variables instead of just one.
Understanding multivariable functions involves several key features:
  • Gradients and Tangents: Derivatives like \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) contribute to calculating the gradient, which points in the direction of greatest change.
  • Dependency of Variables: Here, all variables are interdependent. In practice, modifying one variable (like \(x\)) might impact another (like \(z\)), indirectly affecting \(y\) as well.
This interconnectedness highlights the complexity and beauty of multivariable calculus. It’s crucial in fields such as engineering and computer graphics, where systems involve multiple, interdependent variables.

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Most popular questions from this chapter

A rectangular building is being designed to minimize heat loss. The east and west walls lose heat at a rate of 10 units/m \(^{2}\) per day, the north and south walls at a rate of 8 units/m \(^{2}\) per day, the floor at a rate of 1 unit/m \(^{2}\) per day, and the roof at a rate of 5 units/m \(^{2}\) per day. Each wall must be at least 30 \(\mathrm{m}\) long, the height must be at least \(4 \mathrm{m},\) and the volume must be exactly 4000 \(\mathrm{m}^{3}\) . (a) Find and sketch the domain of the heat loss as a function of the lengths of the sides. (b) Find the dimensions that minimize heat loss. (Check both the critical points and the points on the boundary of the domain.) (c) Could you design a building with even less heat loss if the restrictions on the lengths of the walls were removed?

Find the maximum and minimum values of \(f\) subject to the given constraints. Use a computer algebra system to solve the system of equations that arises in using Lagrange multipliers. (If your CAS finds only one solution, you may need to use additional commands.) $$ f(x, y, z)=x+y+z ; \quad x^{2}-y^{2}=z, x^{2}+z^{2}=4 $$

Suppose \((1,1)\) is a critical point of a function \(f\) with continuous second derivatives. In each case, what can you say about \(f ?\) $$\begin{array}{ll}{\text { (a) } f_{x x}(1,1)=4,} & {f_{x y}(1,1)=1, \quad f_{y y}(1,1)=2} \\ {\text { (b) } f_{x x}(1,1)=4,} & {f_{x y}(1,1)=3, \quad f_{y y}(1,1)=2}\end{array}$$

(a) Maximize \(\sum_{i=1}^{n} x_{i} y_{i}\) subject to the constraints \(\sum_{i=1}^{n} x_{i}^{2}=1\) and \(\sum_{i=1}^{n} y_{i}^{2}=1\) (b) Put $$ x_{i}=\frac{a_{i}}{\sqrt{\sum a_{j}^{2}}} \quad \text { and } \quad y_{i}=\frac{b_{i}}{\sqrt{\sum b_{j}^{2}}} $$ to show that $$ \sum a_{i} b_{i} \leqslant \sqrt{\sum a_{j}^{2}} \sqrt{\Sigma b_{j}^{2}} $$ for any numbers \(a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n} .\) This inequality is known as the Cauchy-Schwarz Inequality.

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=x y-2 x-2 y-x^{2}-y^{2}$$

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