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Magazine Chapter 14: Problem 6
Find \(d z / d t\) using the chain rule. Assume the variables are restricted to domains on which the functions are defined. $$z=(x+y) e^{y}, x=2 t, y=1-t^{2}$$
Short Answer
Expert verified
\( \frac{dz}{dt} = e^{1-t^2} [2 - 2t(2t + 2 - t^2)] \).
Step by step solution
01
Identify Variables and Functions
First, identify the primary function and the dependent variables. Here, the function is \( z = (x+y) e^y \). The variables \( x = 2t \) and \( y = 1 - t^2 \) are given explicitly as functions of \( t \).
02
Differentiate \( x \) and \( y \) with respect to \( t \)
Differentiate the given expressions for \( x \) and \( y \) with respect to \( t \). This gives us:\( \frac{d x}{d t} = 2 \) and \( \frac{d y}{d t} = -2t \).
03
Use the Chain Rule to Differentiate \( z \) with Respect to \( t \)
Apply the chain rule for differentiation. The chain rule states that:\( \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \).We need to find the partial derivatives \( \frac{\partial z}{\partial x} \) and \(\frac{\partial z}{\partial y} \).
04
Find \( \frac{\partial z}{\partial x} \)
Differentiate \( z = (x+y)e^y \) with respect to \( x \), treating \( y \) as a constant:\( \frac{\partial z}{\partial x} = e^y \).
05
Find \( \frac{\partial z}{\partial y} \)
Differentiate \( z = (x+y)e^y \) with respect to \( y \):Using the product rule,\( \frac{\partial (x+y)}{\partial y} = 1 \) and \( \frac{\partial}{\partial y}(e^y) = e^y \).Thus, \( \frac{\partial z}{\partial y} = (x+y)e^y + (x+y)e^y = (x+y+1)e^y \).
06
Substitute Partial Derivatives Into the Chain Rule
Now substitute these partial derivatives and the derivatives of \( x \) and \( y \) into the chain rule expression:\( \frac{dz}{dt} = e^y \cdot 2 + (x+y+1)e^y \cdot (-2t) \).
07
Substitute Back \( x \) and \( y \) in the Expression
Substitute \( x = 2t \) and \( y = 1-t^2 \) back into the expression:\[ z = (2t + 1 - t^2)e^{1-t^2} \].Thus:\( \frac{dz}{dt} = e^{1-t^2} \cdot 2 + (2t + 1 - t^2 + 1)e^{1-t^2} \cdot (-2t) \).
08
Simplify the Expression
Simplify the expression for \( \frac{dz}{dt} \):\( \frac{dz}{dt} = 2e^{1-t^2} - 2t(2t + 2 - t^2)e^{1-t^2} \).Further simplifying:\[ \frac{dz}{dt} = e^{1-t^2} [2 - 2t(2t + 2 - t^2)] \].
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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 function changes as one of its variables changes while keeping others constant. It鈥檚 like a peek at how each separate direction influences the function.
In our exercise, we dealt with the function \( z = (x+y) e^y \). We needed to find partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) to apply the Chain Rule effectively.
Here鈥檚 how it worked:
In our exercise, we dealt with the function \( z = (x+y) e^y \). We needed to find partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) to apply the Chain Rule effectively.
Here鈥檚 how it worked:
- To find \( \frac{\partial z}{\partial x} \), we treated \( y \) as a constant. This simplifies our function, making differentiation simpler. For \( z = (x+y) e^y \), the derivative with respect to \( x \) gives \( e^y \), since \( x \) is linear with respect to \( z \).
- For \( \frac{\partial z}{\partial y} \), both \( (x+y) \) and \( e^y \) change with \( y \), requiring the product rule. The result was \( (x + y + 1) e^y \), showing the combined change due to both parts.
Differentiation
Differentiation is the mathematical technique that involves finding the derivative, or the rate of change of a function with respect to a variable. It forms the backbone of understanding how functions behave when their inputs change.
In our scenario, the task included differentiating specific expressions like \( x = 2t \) and \( y = 1 - t^2 \) with respect to \( t \).
Here's the breakdown:
In our scenario, the task included differentiating specific expressions like \( x = 2t \) and \( y = 1 - t^2 \) with respect to \( t \).
Here's the breakdown:
- By differentiating \( x = 2t \) with respect to \( t \), we obtain \( \frac{dx}{dt} = 2 \). This tells us that \( x \) increases at a steady rate as \( t \) increases.
- Similarly, differentiating \( y = 1 - t^2 \) with respect to \( t \) yields \( \frac{dy}{dt} = -2t \), which captures how \( y \) decreases as \( t \) squares. The negative sign indicates a decrease when \( t \) is positive.
Function of a Function
The concept of a "function of a function" is foundational in calculus and appears prominently with the chain rule. We call these compositions, where one function takes another function as its input.
In this exercise, \( z = (x+y)e^y \), where \( x = 2t \) and \( y = 1 - t^2 \), is an example of a function of a function. Here, \( z \) depends on \( t \) indirectly through its dependence on \( x \) and \( y \).
Let鈥檚 consider this structure:
In this exercise, \( z = (x+y)e^y \), where \( x = 2t \) and \( y = 1 - t^2 \), is an example of a function of a function. Here, \( z \) depends on \( t \) indirectly through its dependence on \( x \) and \( y \).
Let鈥檚 consider this structure:
- The primary function, \( z \), relies on \( x+y \) as its input, which further relies on the variable \( t \).
- Every change in \( t \) influences \( x \) and \( y \), thus impacting \( z \) through the relationship defined by the expressions.
- To navigate such relationships, the chain rule provides a pathway to link the derivatives of these nested functions to show how each layer interacts with \( t \).
