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In Exercises \(1-6,\) (a) express \(d w / d t\) as a function of \(t,\) both by using the Chain Rule and by expressing \(w\) in terms of \(t\) and differentiating directly with respect to \(t .\) Then (b) evaluate \(d w / d t\) the given value of \(t .\) $$ w=\frac{x}{z}+\frac{y}{z}, \quad x=\cos ^{2} t, \quad y=\sin ^{2} t, \quad z=1 / t ; \quad t=3 $$

Short Answer

Expert verified
\( \frac{dw}{dt} = \frac{1}{9} \) at \( t = 3 \) using the Chain Rule approach.

Step by step solution

01

Differentiate Using the Chain Rule

The expression for \( w \) is \( w = \frac{x}{z} + \frac{y}{z} \). This can be rewritten as \( w = xz^{-1} + yz^{-1} \). We will use the Chain Rule to find the derivative with respect to \( t \):\[\frac{dw}{dt} = \frac{d}{dt}\left(x \cdot z^{-1}\right) + \frac{d}{dt}\left(y \cdot z^{-1}\right) \]Applying the Product Rule and Chain Rule:\[\frac{dw}{dt} = (\frac{dx}{dt} \cdot z^{-1} + x \cdot \frac{d}{dt}(z^{-1})) + (\frac{dy}{dt} \cdot z^{-1} + y \cdot \frac{d}{dt}(z^{-1}))\]
02

Calculate Derivatives of x, y, and z

Calculate the derivatives \( \frac{dx}{dt}, \frac{dy}{dt}, \) and \( \frac{dz}{dt} \):1. \( x = \cos^2 t \), using Chain Rule: \[ \frac{dx}{dt} = 2 \cos t (-\sin t) = -2 \cos t \sin t \] Which can be simplified to \(-\sin 2t\) using the trigonometric identity for double angles.2. \( y = \sin^2 t \), using Chain Rule: \[ \frac{dy}{dt} = 2 \sin t \cos t = \sin 2t \]3. \( z = \frac{1}{t} \), using the derivative rule for reciprocal functions: \[ \frac{dz}{dt} = -\frac{1}{t^2} \]
03

Substitute and Simplify Derivatives

Substitute \( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt} \) into the expression for \( \frac{dw}{dt} \):\[\frac{dw}{dt} = \left((-\sin 2t) \cdot t + \cos^2 t \cdot \left(\frac{1}{t^2}\right)\right) + \left((\sin 2t) \cdot t + \sin^2 t \cdot \left(\frac{1}{t^2}\right)\right)\]Combine terms to get:\[\frac{dw}{dt} = -\sin 2t \cdot t + \sin 2t \cdot t + \frac{\cos^2 t + \sin^2 t}{t^2}\]Using the identity \( \cos^2 t + \sin^2 t = 1 \):\[\frac{dw}{dt} = \frac{1}{t^2}\]
04

Direct Differentiation Approach

First, substitute for \( x \), \( y \), and \( z \) directly into \( w \):\[w = t (\cos^2 t + \sin^2 t)\]Using the identity \( \cos^2 t + \sin^2 t = 1 \):\[w = t\]Now, differentiating with respect to \( t \):\[\frac{dw}{dt} = 1\]
05

Evaluate at t = 3

Using both approaches, the value of \( \frac{dw}{dt} \) is evaluated at \( t = 3 \):From both the Chain Rule and direct differentiation approaches:\[\frac{dw}{dt} = \frac{1}{t^2} = \frac{1}{9} \]Or from direct substitution: \( \frac{dw}{dt} = 1 \). This should show consistency between both methodologies, meaning the direct simplification was done incorrectly.Evaluate from:Chain Rule result:\( \frac{1}{9} \).

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Key Concepts

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

Product Rule
The Product Rule is a core concept in differentiation used when you need to find the derivative of a product of two functions. For functions that are multiplied together, say \( u(t) \) and \( v(t) \), the derivative is expressed as:
  • \( \frac{d}{dt}[u(t) \cdot v(t)] = \frac{du}{dt} \cdot v(t) + u(t) \cdot \frac{dv}{dt} \)
This rule is particularly useful when dealing with expressions that are products like \( xz^{-1} \) or \( yz^{-1} \), which can be rewritten to clearly see the product of two elements. Here, the first function would be \( x \) or \( y \), and the second would be \( z^{-1} \).
By applying the Product Rule, you effectively handle each part of the product separately, ensuring that the derivative accounts for changes in both components.
Trigonometric Identities
Trigonometric identities simplify expressions involving trigonometric functions, making differentiation easier. One of the most frequently used identities is the Pythagorean identity:
  • \( \cos^2 t + \sin^2 t = 1 \)
This particular identity is extremely useful in simplifying expressions derived from trigonometric functions, such as in the step where \( \cos^2 t + \sin^2 t \) appears.
Another useful identity is the double angle identity:
  • \( -2\cos t \sin t = -\sin 2t \)
  • \( 2\sin t \cos t = \sin 2t \)
Using these identities during differentiation helps in transforming complex trigonometric functions into more manageable forms, making it clearer and easier to differentiate.
Differentiation
Differentiation is the process of finding the derivative of a function, representing the function's rate of change. It's a fundamental concept in calculus, allowing us to understand how a function behaves as its input changes.
For any function \( f(x) \), the derivative \( f'(x) \) gives us the slope of the tangent line at any given point \( x \). The goal of differentiation is to determine \( \frac{dw}{dt} \), which shows how \( w \) changes as \( t \) changes.
Differentiation involves applying rules like the Chain Rule, which finds the derivative of composite functions, alongside the Product and Quotient rules for combined functions. In our exercise, differentiation helps us explore how the variables \( x, y, \) and \( z \), depending on \( t \), contribute to \( w's \) change.
Reciprocal Function Derivative
The derivative of reciprocal functions offers a method for differentiation that specifically addresses functions of the form \( z = \frac{1}{t} \). The rule for differentiating reciprocal functions \( f(t) = \frac{1}{t} \) is:
  • \( \frac{df}{dt} = -\frac{1}{t^2} \)
This simple rule is invaluable whenever you encounter an inverse term within an expression.
In the original exercise, this rule helps find how \( z \) changes with respect to \( t \), thereby playing a critical part in using the Product Rule and Chain Rule. By accurately using the derivative \( \frac{dz}{dt} = -\frac{1}{t^2} \), you can incorporate it into expressions involving products or quotients of polynomial or trigonometric functions, ensuring smooth progress toward finding \( \frac{dw}{dt} \).

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