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Magazine Chapter 14: Problem 5
Find \(d z / d t\) using the chain rule. Assume the variables are restricted to domains on which the functions are defined. $$z=x e^{y}, x=2 t, y=1-t^{2}$$
Short Answer
Expert verified
\(\frac{dz}{dt} = (2 - 4t^2) e^{(1-t^2)}\)
Step by step solution
01
Understand the Problem
We need to find \(\frac{dz}{dt}\), where \(z = x e^y\) is a function of \(t\) via \(x = 2t\) and \(y = 1 - t^2\). This requires using the chain rule.
02
Express z as a Function of t
Substitute \(x = 2t\) and \(y = 1 - t^2\) into \(z = xe^y\):\[ z = (2t) e^{(1-t^2)} \] This shows that \(z\) is explicitly a function of \(t\).
03
Differentiate z with Respect to t
We use the product rule on \(z = 2t e^{(1-t^2)}\), where the first function is \(u = 2t\) and the second is \(v = e^{(1-t^2)}\). The derivative \(\frac{dz}{dt} = \frac{du}{dt}v + u\frac{dv}{dt}\).
04
Derivative of u (2t) with Respect to t
Calculate \(\frac{du}{dt}\): Since \(u = 2t\), then \(\frac{du}{dt} = 2\).
05
Derivative of v \(e^{(1-t^2)}\) with Respect to t
Calculate \(\frac{dv}{dt}\) using the chain rule: \[ \frac{dv}{dt} = \frac{d}{dt} \left( e^{1-t^2} \right) = e^{1-t^2} \cdot (-2t) \] After simplifying, \(\frac{dv}{dt} = -2te^{(1-t^2)}\).
06
Apply the Product Rule
Substitute \(\frac{du}{dt}\) and \(\frac{dv}{dt}\) into the product rule formula:\[ \frac{dz}{dt} = 2 imes e^{(1-t^2)} + 2t imes (-2t e^{(1-t^2)}) \] Simplify this expression: \[ \frac{dz}{dt} = 2e^{(1-t^2)} - 4t^2 e^{(1-t^2)} \] Combine terms: \[ \frac{dz}{dt} = (2 - 4t^2) e^{(1-t^2)} \]
07
Final Expression for dz/dt
The solution for \(\frac{dz}{dt}\) is: \[ \frac{dz}{dt} = (2 - 4t^2) e^{(1-t^2)} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule for Differentiation
The product rule is a handy tool in calculus when you need to differentiate a function that is the product of two or more functions. Essentially, it's used to differentiate expressions like our function \(z = 2t e^{(1-t^2)}\). Here, you can break down the function into two parts: \(u = 2t\) and \(v = e^{1-t^2}\).
- According to the product rule, if you have a function \(z = uv\), the derivative \(\frac{dz}{dt}\) is given by \(\frac{du}{dt}v + u\frac{dv}{dt}\).
- This means you find the derivative of \(u\) with respect to \(t\) (\(\frac{du}{dt}\)), and multiply it by \(v\), then add \(u\) multiplied by the derivative of \(v\) with respect to \(t\) (\(\frac{dv}{dt}\)).
Differentiation Basics
Differentiation is a central concept in calculus that involves finding the rate at which a function changes at any given point. It's the process of computing a derivative, which tells us the slope of a function's graph at a certain point. In practical terms, differentiation lets us understand how one variable changes concerning another.Here are some simple points about differentiation:
- Given a function \(f(t)\), its derivative \(f'(t)\) represents the rate of change of \(f\) with respect to \(t\).
- For example, if \(f(t) = 2t\), then the differentiation yields \(f'(t) = 2\), indicating a constant rate of change.
Exponential Function and Its Differentiation
The exponential function, denoted generally as \(e^x\), is a critical function in mathematics due to its unique properties, especially when it comes to differentiation. For an exponential term like \(e^{(1-t^2)}\), understanding how to differentiate it helps unlock numerous powerful calculus applications.Some interesting properties of exponential functions include:
- The derivative of \(e^x\) is \(e^x\) itself, showcasing the exponential growth property.
- When differentiating \(e^{g(t)}\), where \(g(t)\) is any function of \(t\), you must apply the chain rule: \(\frac{d}{dt} (e^{g(t)}) = e^{g(t)} \cdot g'(t)\).
