Problem 227
Question
Let \(w(x, y, z)=x^{2}+y^{2}+z^{2}\) \(x=\cos t, y=\sin t\) and \(z=e^{t}\) Express \(w\) as a function of \(t\) and find \(\frac{d w}{d t}\) directly. Then, find \(\frac{d w}{d t}\) using the chain rule.
Step-by-Step Solution
Verified Answer
\(\frac{d w}{d t} = 2e^{2t}\) using both methods.
1Step 1: Substitute Parametric Equations
To express \(w\) as a function of \(t\), substitute the given parametric equations into the expression for \(w\). This gives us \(w(t) = \cos^2 t + \sin^2 t + (e^t)^2\).
2Step 2: Simplify the Function
Using the Pythagorean identity \(\cos^2 t + \sin^2 t = 1\), simplify the expression to \(w(t) = 1 + e^{2t}\).
3Step 3: Differentiate Directly
Differentiate \(w(t) = 1 + e^{2t}\) directly with respect to \(t\). The derivative is \(\frac{d w}{d t} = 2e^{2t}\).
4Step 4: Find Partial Derivatives for the Chain Rule
Calculate the partial derivatives: \(\frac{\partial w}{\partial x} = 2x\), \(\frac{\partial w}{\partial y} = 2y\), and \(\frac{\partial w}{\partial z} = 2z\).
5Step 5: Differentiate Parametric Equations
Differentiate the parametric equations with respect to \(t\): \(\frac{d x}{d t} = -\sin t\), \(\frac{d y}{d t} = \cos t\), and \(\frac{d z}{d t} = e^t\).
6Step 6: Apply the Chain Rule
Apply the chain rule: \(\frac{d w}{d t} = \frac{\partial w}{\partial x}\frac{d x}{d t} + \frac{\partial w}{\partial y}\frac{d y}{d t} + \frac{\partial w}{\partial z}\frac{d z}{d t}\). Plug in the values to get \(2x(-\sin t) + 2y(\cos t) + 2z(e^t)\), which simplifies to the same result, \(2e^{2t}\).
7Step 7: Confirm the Result
Verify both methods give the same outcome: \(\frac{d w}{d t} = 2e^{2t}\), confirming correctness.
Key Concepts
Understanding Parametric EquationsExploring Partial Derivatives and Their RoleDifferentiation and the Chain Rule
Understanding Parametric Equations
Parametric equations are mathematical expressions where a set of equations express the coordinates of the points of a geometric object as functions of a variable, often time (denoted as \( t \)).
This technique is useful in tracing paths of objects that do not follow traditional linear equations. In our exercise, the parametric equations are \( x = \cos t \), \( y = \sin t \), and \( z = e^{t} \).
Each equation assigns values based on the parameter \( t \), allowing us to derive relationships and properties about the object.
This technique is useful in tracing paths of objects that do not follow traditional linear equations. In our exercise, the parametric equations are \( x = \cos t \), \( y = \sin t \), and \( z = e^{t} \).
Each equation assigns values based on the parameter \( t \), allowing us to derive relationships and properties about the object.
- These equations help simplify complex movements into a single variable system.
- They are commonly used in physics, engineering, and computer graphics.
- From our exercise, substituting these in \( w = x^2 + y^2 + z^2 \) builds a new function \( w(t) \).
Exploring Partial Derivatives and Their Role
Partial derivatives are derivatives of multivariable functions with respect to one variable, keeping the others constant.
In the context of our exercise, \( w(x, y, z) \) has partial derivatives \( \frac{\partial w}{\partial x} = 2x \), \( \frac{\partial w}{\partial y} = 2y \), and \( \frac{\partial w}{\partial z} = 2z \).
These derivatives represent the rate of change of \( w \) with respect to each variable independently.
In the context of our exercise, \( w(x, y, z) \) has partial derivatives \( \frac{\partial w}{\partial x} = 2x \), \( \frac{\partial w}{\partial y} = 2y \), and \( \frac{\partial w}{\partial z} = 2z \).
These derivatives represent the rate of change of \( w \) with respect to each variable independently.
- Calculating partial derivatives is a fundamental step in applying the chain rule.
- They lay the foundation for finding the total derivative in the case of parametric equations.
- By analyzing these, one can understand how changes in each variable impact the overall function \( w \).
Differentiation and the Chain Rule
Differentiation is the process of finding a derivative, which provides the slope or rate of change of a function.
In a parametric context, like in our exercise, differentiation can be approached directly or via the chain rule.
The derivative \( \frac{d w}{d t} \) is calculated directly as \( 2e^{2t} \) or using the chain rule for verification.
In a parametric context, like in our exercise, differentiation can be approached directly or via the chain rule.
The derivative \( \frac{d w}{d t} \) is calculated directly as \( 2e^{2t} \) or using the chain rule for verification.
- The chain rule combines partial derivatives with derivatives of the parameter equations.
- It offers a way to find derivatives when the function is composed of other functions, as seen with \( w(x,y,z) \).
- Using the chain rule, \( \frac{d w}{d t} \) is derived by summing the products of partial derivatives and the rates of each corresponding variable with respect to \( t \).
Other exercises in this chapter
Problem 225
For the following exercises, find \(\frac{d f}{d t}\) using the chain rule and direct substitution. $$ f(x, y)=\ln (x+y), \quad x=e^{t}, y=e^{t} $$
View solution Problem 226
For the following exercises, find \(\frac{d f}{d t}\) using the chain rule and direct substitution. $$ f(x, y)=x^{4}, \quad x=t, y=t $$
View solution Problem 228
Let \(z=x^{2} y, \quad\) where \(x=t^{2}\) and \(y=t^{3} .\) Find \(\frac{d z}{d t}\)
View solution Problem 229
Let \(u=e^{x} \sin y\) where \(x=t^{2}\) and \(y=\pi t\). Find \(\frac{d u}{d t}\) when \(x=\ln 2\) and \(y=\frac{\pi}{4}\)
View solution