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Magazine Chapter 14: Problem 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 \(d w / d t\) at the given value of \(t\). \(\begin{array}{l}w=z-\sin x y, \quad x=t, \quad y=\ln t, \quad z=e^{t-1} ; \quad t=1\end{array}\)
Short Answer
Expert verified
\(\frac{dw}{dt} = 0\) at \(t = 1\).
Step by step solution
01
Set Up the Problem
We have a function \(w = z - \sin(xy)\), where \(x = t\), \(y = \ln t\), and \(z = e^{t-1}\). Our task is to find \(\frac{dw}{dt}\) using the chain rule and by direct substitution.
02
Use the Chain Rule
Using the chain rule, we write \(\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}\). Let's find each term. \(\frac{\partial w}{\partial x} = -y \cos(xy)\), and since \(x = t\), we have \(\frac{dx}{dt} = 1\). \(\frac{\partial w}{\partial y} = -x \cos(xy)\), and \(\frac{dy}{dt} = \frac{1}{t}\). \(\frac{\partial w}{\partial z} = 1\), and \(\frac{dz}{dt} = e^{t-1}\). Substitute these into the chain rule equation.
03
Substitute Chain Rule Components
Substituting, we get: \[\frac{dw}{dt} = (-y \cos(xy))(1) + (-x \cos(xy))\left(\frac{1}{t}\right) + (1)(e^{t-1})\]. Simplifying, \(\frac{dw}{dt} = -y \cos(xy) - \frac{x}{t} \cos(xy) + e^{t-1}\).
04
Express w in terms of t Directly
Now, replace \(x\), \(y\), and \(z\) with \(t\)-expressions in \(w = z - \sin(xy)\). We get \(w = e^{t-1} - \sin(t \ln t)\). Differentiating directly, \(\frac{dw}{dt} = \frac{d}{dt}[e^{t-1}] - \frac{d}{dt}[\sin(t \ln t)]\).
05
Differentiate Each Part
The derivative of \(e^{t-1}\) with respect to \(t\) is \(e^{t-1}\). Using the product rule for \(\sin(t \ln t)\), we have \(\frac{d}{dt}[t \ln t] = 1 \cdot \ln t + t \cdot \frac{1}{t} = \ln t + 1\). Thus, \(\frac{d}{dt}[\sin(t \ln t)] = \cos(t \ln t)(\ln t + 1)\). Combine for \(\frac{dw}{dt} = e^{t-1} - \cos(t \ln t)(\ln t + 1)\).
06
Evaluate at t = 1
For \(t=1\), we substitute into both expressions. Using the chain rule expression: \(y = \ln 1 = 0\), so \(\cos(xy) = 1\); thus, the term becomes \(-0 - \frac{1}{1} \cdot 1 + e^{0} = 1 - 1 = 0\). Direct approach gives the same: \(\ln 1 = 0\), \(\sin(0) = 0\), so \(e^{0} - 0 = 1\). Then differentiate and evaluate: \(e^{0} - \cos(0)(0+1) = 1 - 1 = 0\).
07
Conclude
Both methods yield \(\frac{dw}{dt} = 0\) at \(t = 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives help us understand how a multivariable function changes as one of the input variables changes, while keeping others constant. In this exercise, we have a function \( w = z - \sin(xy) \), where \( x \), \( y \), and \( z \) are all functions of \( t \). To find how \( w \) changes as \( t \) changes, we use partial derivatives to focus on one variable at a time.
Here's how it works:
Here's how it works:
- \( \frac{\partial w}{\partial x} \): This tells us how \( w \) changes when \( x \) changes, while \( y \) and \( z \) are fixed. For our function, \( \frac{\partial w}{\partial x} = -y \cos(xy) \), showing the rate of change is dependent on both \( y \) and the cosine of \( xy \).
- \( \frac{\partial w}{\partial y} \): This indicates how \( w \) changes with \( y \), with \( x \) and \( z \) held constant. Here, \( \frac{\partial w}{\partial y} = -x \cos(xy) \).
- \( \frac{\partial w}{\partial z} \): Since \( z \) is a simple variable subtraction, its partial derivative is \( 1 \).
Product Rule in Calculus
The product rule is used when differentiating expressions that are the product of two functions. It's particularly handy when functions are multiplied together, as seen in our expression \( \sin(t \ln t) \) when we differentiated \( t \ln t \) as part of \( w \).
To apply the product rule, remember:
To apply the product rule, remember:
- If \( u(t) \) and \( v(t) \) are functions of \( t \), the product rule is \( \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \).
- In our case, set \( u(t) = t \) and \( v(t) = \ln t \). So, \( \frac{d}{dt}[t \ln t] = 1 \times \ln t + t \times \frac{1}{t} = \ln t + 1 \).
Function Differentiation
Function differentiation is like peeling an onion, layer by layer, to understand the rate of change within a function. It allows us to see how a function responds to changes in its input variables鈥攁 crucial skill in calculus.
In the context of our exercise, differentiation involves both chain and product rule tactics:
In the context of our exercise, differentiation involves both chain and product rule tactics:
- The direct differentiation of \( w \) involves determining how each component, notably \( e^{t-1} \) and \( \sin(t \ln t) \), responds to changes in \( t \).
- When differentiating \( e^{t-1} \), recognize that it is a straightforward exponential function. Its derivative, consistent with exponential rules, is \( e^{t-1} \).
- For \( \sin(t \ln t) \), differentiate the inner function first using the product rule, then employ the outer derivative \( \cos(t \ln t) \).
