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Use the chain rule to find \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t},\) where $$z=e^{x y} \tan y, x=4 s+5 t, y=\frac{3 s}{3 t}$$ First the pieces: \(\frac{\partial z}{\partial x}=\) ________ \(\frac{\partial z}{\partial y}=\) \(\frac{\partial x}{\partial s}=\) ________ \(\frac{\partial x}{\partial t}=\) \(\frac{\partial y}{\partial s}=\) ________ \(\frac{\partial y}{\partial t}=\)

Short Answer

Expert verified
\( \frac{\partial z}{\partial x} = e^{(xy)}\tan{y}\cdot y \) \( \frac{\partial z}{\partial y} = e^{(xy)}\sec^2{y}\cdot x + e^{(xy)}\tan{y} \) \( \frac{\partial x}{\partial s} = 4 \) \( \frac{\partial x}{\partial t} = 5 \) \( \frac{\partial y}{\partial s} = \frac{3}{3t} \) \( \frac{\partial y}{\partial t} = -\frac{3s}{(3t)^2} \) \( \frac{\partial{z}}{\partial{s}} = (e^{(xy)}\tan{y}\cdot y)(4) + (e^{(xy)}\sec^2{y}\cdot x + e^{(xy)}\tan{y})(\frac{3}{3t}) \\ \) \( \frac{\partial{z}}{\partial{t}} = (e^{(xy)}\tan{y}\cdot y)(5) + (e^{(xy)}\sec^2{y}\cdot x + e^{(xy)}\tan{y})(-\frac{3s}{(3t)^2}) \\ \)

Step by step solution

01

Find the partial derivatives of z with respect to x and y

Given: \(z = e^{(xy)} \tan{y}\), Now find the partial derivatives of z with respect to x and y, i.e, \(\frac{\partial{z}}{\partial{x}}\) and \(\frac{\partial{z}}{\partial{y}}\). \(\frac{\partial{z}}{\partial{x}} = e^{(xy)}\tan{y}\cdot y\) \(\frac{\partial{z}}{\partial{y}} = e^{(xy)}\sec^2{y}\cdot x + e^{(xy)}\tan{y}\)
02

Find the partial derivatives of x and y with respect to s and t

Given: \(x = 4s + 5t\), \(y = \frac{3s}{3t}\), Now find the partial derivatives of x and y with respect to s and t, i.e, \(\frac{\partial{x}}{\partial{s}}\), \(\frac{\partial{x}}{\partial{t}}\), \(\frac{\partial{y}}{\partial{s}}\), and \(\frac{\partial{y}}{\partial{t}}\). \(\frac{\partial{x}}{\partial{s}} = 4\) \(\frac{\partial{x}}{\partial{t}} = 5\) \(\frac{\partial{y}}{\partial{s}} = \frac{3}{3t}\) \(\frac{\partial{y}}{\partial{t}} = -\frac{3s}{(3t)^2}\)
03

Apply chain rule to find the partial derivatives of z with respect to s and t

Now apply the chain rule to find \(\frac{\partial{z}}{\partial{s}}\) and \(\frac{\partial{z}}{\partial{t}}\). \( \frac{\partial{z}}{\partial{s}} = \frac{\partial{z}}{\partial{x}}\frac{\partial{x}}{\partial{s}} + \frac{\partial{z}}{\partial{y}}\frac{\partial{y}}{\partial{s}} \\ \) \( \frac{\partial{z}}{\partial{s}} = (e^{(xy)}\tan{y}\cdot y)(4) + (e^{(xy)}\sec^2{y}\cdot x + e^{(xy)}\tan{y})(\frac{3}{3t}) \\ \) Similarly, we have \( \frac{\partial{z}}{\partial{t}} = \frac{\partial{z}}{\partial{x}}\frac{\partial{x}}{\partial{t}} + \frac{\partial{z}}{\partial{y}}\frac{\partial{y}}{\partial{t}} \\ \) \( \frac{\partial{z}}{\partial{t}} = (e^{(xy)}\tan{y}\cdot y)(5) + (e^{(xy)}\sec^2{y}\cdot x + e^{(xy)}\tan{y})(-\frac{3s}{(3t)^2}) \\ \)

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Key Concepts

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

Partial Derivatives
Partial derivatives are essential tools in multivariable calculus. They allow us to understand how a function changes as one of its variables changes, while all other variables are held constant. For a function of multiple variables, like \( z = e^{xy} \tan y \), a partial derivative with respect to \( x \) would measure the rate of change of \( z \) when there is a small change in \( x \), keeping \( y \) constant.

This concept is crucial because it gives insight into the slope of the function in the direction of the variable of interest. In our calculation:
  • \( \frac{\partial z}{\partial x} = e^{(xy)} \tan{y} \cdot y \) tells us how \( z \) changes with \( x \).
  • \( \frac{\partial z}{\partial y} = e^{(xy)} \sec^2{y} \cdot x + e^{(xy)} \tan{y} \) indicates the change in \( z \) with \( y \).
Exploring these partial derivatives helps us understand the behavior of the function in a multidimensional space, which is the core of multivariable calculus.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions with more than one variable. It deals with deriving and analyzing formulas that involve multiple variables. Common processes include differentiation, integration, and optimization in higher dimensions.

One of the major challenges in multivariable calculus is understanding how multiple variables collectively influence changes in a function. For instance, in our study with \( x = 4s + 5t \) and \( y = \frac{3s}{3t} \), each variable \( s \) and \( t \) affects both \( x \) and \( y \), making it complex to see how they change \( z \) simultaneously.

By using techniques like partial derivatives, we can study each path of influence independently. Additionally, calculus tools like the chain rule bridge these relationships by explaining how changes in the input variables \( s \) and \( t \) affect the output \( z \). This intricate understanding aids in solving real-world problems involving multiple variables, spanning fields such as physics, engineering, and economics.
Calculus Problem Solving
Solving calculus problems, especially involving multivariable functions, requires careful application of computational rules and a clear understanding of the underlying concepts. The step-by-step solution involves:

- **Identifying Functions and Relationships**: Recognize the primary function, \( z = e^{xy} \tan y \), and rewrite it to see which variables influence one another, expressed here as \( x = 4s + 5t \) and \( y = \frac{3s}{3t} \).
- **Differentiating Using Partial Derivatives**: Determine how \( z \) changes with each of the embedded variables \( x \) and \( y \) by computing \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).
- **Application of the Chain Rule**: Implement the chain rule for derivatives to interconnect the changes from \( s \) and \( t \) to \( z \) via their relationships through \( x \) and \( y \). This involves finding \( \frac{\partial x}{\partial s}, \frac{\partial x}{\partial t}, \frac{\partial y}{\partial s}, \) and \( \frac{\partial y}{\partial t} \), and then combining them through the formula:
  • \( \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} \)
  • \( \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t} \)
This structured approach, breaking complex problems into manageable pieces, allows us to tackle challenging calculus problems successfully.

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Most popular questions from this chapter

Suppose \(f(x, y)=x y-a x-b y\). (A) How many local minimum points does \(f\) have in \(\mathbf{R}^{2} ?\) (The answer is an integer). (B) How many local maximum points does \(f\) have in \(\mathbf{R}^{2} ?\) (C) How many saddle points does \(f\) have in \(\mathbf{R}^{2} ?\)

(a) Check the local linearity of \(f(x, y)=e^{-x} \cos (y)\) near \(x=1, y=1.5\) by filling in the following table of values of \(f\) for \(x=0.9,1,1.1\) and \(y=1.4,1.5,1.6 .\) Express values of \(f\) with 4 digits after the decimal point. (b) Next, fill in the table for the values \(x=0.99,1,1.01\) and \(y=\) 1.49,1.5,1.51 , again showing 4 digits after the decimal point. Notice if the two tables look nearly linear, and whether the second looks more linear than the first (in particular, think about how you would decide if they were linear, or if the one were more closely linear than the other). (c) Give the local linearization of \(f(x, y)=e^{-x} \cos (y)\) at (1,1.5) : Using the second of your tables: \(f(x, y) \approx\) ____________. Using the fact that \(f_{x}(x, y)=-e^{-x} \cos (y)\) and \(f_{y}(x, y)=-e^{-x} \sin (y):\) \(f(x, y) \approx\) ___________.

Suppose \(w=\frac{x}{y}+\frac{y}{z},\) where $$x=e^{5 t}, y=2+\sin (3 t), \text { and } z=2+\cos (6 t)$$ A) Use the chain rule to find \(\frac{d w}{d t}\) as a function of \(x, y, z,\) and \(t .\) Do not rewrite \(x, y,\) and \(z\) in terms of \(t,\) and do not rewrite \(e^{5 t}\) as \(x\). \(\frac{d w}{d t}=\) ____________. B) Use part A to evaluate \(\frac{d w}{\|}\) when \(t=0\).

Find the partial derivatives of the function $$f(x, y)=\int_{y}^{x} \cos \left(7 t^{2}+8 t-2\right) d t$$ \(f_{x}(x, y)=\) ___________. \(f_{y}(x, y)=\) ___________.

The radius of a right circular cone is increasing at a rate of 3 inches per second and its height is decreasing at a rate of 5 inches per second. At what rate is the volume of the cone changing when the radius is 40 inches and the height is 40 inches? _________ cubic inches per second.

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