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Chapter 9: Problem 1

Find dw/dt by (3) or (4) . Check the result by substitution and differentiation, (Show the details.) $$w=\sqrt{x^{2}+y^{2}}, x=e^{2 t}, y=e^{-2 t}$$

Short Answer

Expert verified
\( \frac{dw}{dt} = \frac{2(e^{4t} - e^{-4t})}{\sqrt{e^{4t} + e^{-4t}}} \).

Step by step solution

01

Express w in Terms of t

First, express \( w \) as a function of \( t \). Since \( w = \sqrt{x^2 + y^2} \) and we know \( x = e^{2t} \) and \( y = e^{-2t} \), substitute these into the equation:\[ w = \sqrt{(e^{2t})^2 + (e^{-2t})^2} = \sqrt{e^{4t} + e^{-4t}}. \]
02

Differentiate w with Respect to t

Now differentiate \( w = \sqrt{e^{4t} + e^{-4t}} \) with respect to \( t \). Use the chain rule:- If \( u = e^{4t} + e^{-4t} \), then \( w = \sqrt{u} \).- \( \frac{du}{dt} = 4e^{4t} - 4e^{-4t} \).- \( \frac{dw}{du} = \frac{1}{2\sqrt{u}} \).Thus, \( \frac{dw}{dt} = \frac{du}{dt} \cdot \frac{1}{2\sqrt{u}} = \frac{4e^{4t} - 4e^{-4t}}{2\sqrt{e^{4t} + e^{-4t}}} \).- Simplify to get: \[ \frac{dw}{dt} = \frac{2(e^{4t} - e^{-4t})}{\sqrt{e^{4t} + e^{-4t}}}. \]
03

Verify by Substitution and Differentiation

To check our result, substitute the expression for \( w \) back and differentiate step by step.1. Substitute \( x = e^{2t} \) and \( y = e^{-2t} \): - \( \frac{dx}{dt} = 2e^{2t}, \) - \( \frac{dy}{dt} = -2e^{-2t}. \)2. Use chain rule to express \( \frac{dw}{dt} \) in terms of \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \): - \( \frac{dw}{dt} = \frac{1}{2w} (2x \frac{dx}{dt} + 2y \frac{dy}{dt}) \). - Simplify and find: \( \frac{dw}{dt} = \frac{2(e^{4t} - e^{-4t})}{\sqrt{e^{4t} + e^{-4t}}}. \) - This matches our earlier result, confirming correctness.

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Key Concepts

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

Chain Rule
In differential calculus, the chain rule is a formula used to compute the derivative of a composite function. It is particularly useful when dealing with functions dependent on multiple variables that are themselves functions of another variable.
For example, to find the derivative of \( w = \sqrt{x^2 + y^2} \) with \( x = e^{2t} \) and \( y = e^{-2t} \), we need to use the chain rule.
Here's why the chain rule is powerful:
  • It allows us to break a complex differentiation problem into simpler steps, by calculating derivatives of the inner and outer functions separately.
  • It helps connect different functions smoothly, showcasing the relation between dependent and independent variables.
  • In the given exercise, we applied the chain rule when differentiating \( w \) with respect to \( t \). This was done by first expressing \( w \) as \( \sqrt{u} \) (where \( u = e^{4t} + e^{-4t} \)), and differentiating the layers separately.
This approach simplifies understanding and makes computation easier, even for complicated expressions.
Differentiation
Differentiation is the process of finding the derivative, or rate of change, of a function. It is a fundamental tool in calculus, bringing insight into how functions change at any given point.
In the exercise, we differentiate \( w = \sqrt{e^{4t} + e^{-4t}} \) with respect to \( t \).
Steps to differentiate include:
  • Identifying the function and breaking it down if it is composite. For example, separate \( u = e^{4t} + e^{-4t} \) to find \( \frac{du}{dt} = 4e^{4t} - 4e^{-4t} \).
  • Applying differentiation rules systematically, such as power rule and chain rule, to compute individual derivatives.
  • Combining results: Multiply \( \frac{du}{dt} \) with \( \frac{dw}{du} = \frac{1}{2\sqrt{u}} \) to find \( \frac{dw}{dt} \), which gives the rate of change of \( w \) concerning \( t \).
Through differentiation, we gain insights into the behavior of functions, enabling us to predict future or past outcomes effectively.
Function of a Variable
Understanding functions involving variables is crucial in calculus. A function relates an input value to a single output value and shows how quantities change relative to each other.
In the given problem, \( w \) depends on \( x \) and \( y \), each of which is a function of \( t \).
Considerations when dealing with functions of a variable include:
  • Recognizing how changes in one variable (\( t \)) influence others (\( x \), \( y \), and consequently \( w \)).
  • Substituting known functions into others to express them in terms of a single variable, simplifying differentiation.
  • In this context, knowing \( x = e^{2t} \) and \( y = e^{-2t} \) helps in expressing \( w = \sqrt{x^2 + y^2} \) directly in terms of \( t \).
This simplification is not just about ease of computation but also about understanding dynamics and predictive modeling of real-world systems, all through the lens of calculus.

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