Chapter 13: Problem 5
Use an appropriate form of the chain rule to find $d z / d t $$ z=e^{1-x y} ; x=t^{1 / 3}, y=t^{3} $$
Short Answer
Expert verified
\( \frac{dz}{dt} = -e^{1-t^{10/3}} \left( \frac{t^{7/3}}{3} + 3t^{7/3} \right) \).
Step by step solution
01
Understand the Problem
We need to find the derivative \(\frac{dz}{dt}\), where \(z=e^{1-xy}\) and \(x=t^{1/3}\), \(y=t^3\). This involves using the chain rule since \(z\) is a function of \(x\) and \(y\), which are both functions of \(t\).
02
Differentiate with respect to t using the Chain Rule
Apply the chain rule: \(\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}\). This requires finding the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\), as well as the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
03
Compute Partial Derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\)
Firstly, the expression for \(z\) is \(z = e^{1-xy}\). The partial derivative with respect to \(x\) is: \(\frac{\partial z}{\partial x} = -y \cdot e^{1-xy}\). Similarly, the partial derivative with respect to \(y\) is: \(\frac{\partial z}{\partial y} = -x \cdot e^{1-xy}\).
04
Compute Derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
For \(x = t^{1/3}\), the derivative is \(\frac{dx}{dt} = \frac{1}{3}t^{-2/3}\). For \(y = t^3\), the derivative is \(\frac{dy}{dt} = 3t^2\).
05
Calculate \(\frac{dz}{dt}\)
Substitute the partial derivatives and the derivatives calculated into the chain rule formula: \[\frac{dz}{dt} = \left(-y \cdot e^{1-xy}\right) \cdot \frac{1}{3}t^{-2/3} + \left(-x \cdot e^{1-xy}\right) \cdot 3t^2\]Now substitute \(x = t^{1/3}\) and \(y = t^3\) into the equation. Thus, \(\frac{dz}{dt} = \left(-t^3 \cdot e^{1-t^{1/3}t^3}\right) \cdot \frac{1}{3}t^{-2/3} + \left(-t^{1/3} \cdot e^{1-t^{1/3}t^3}\right) \cdot 3t^2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a crucial concept when dealing with functions of multiple variables. Imagine a function that depends on two or more variables, like our function \( z = e^{1-xy} \) that depends on both \( x \) and \( y \). Instead of finding the effect on \( z \) due to a change in all variables at once, partial derivatives allow us to see how changing one variable, while keeping others constant, affects the function.
In practice, this is like asking: "What happens to \( z \) if only \( x \) changes while \( y \) stays the same?" or vice versa. For our function, we calculated the partial derivative with respect to \( x \), \( \frac{\partial z}{\partial x} \), resulting in \( -y\cdot e^{1-xy} \). This shows that for a small change in \( x \), \( z \) will change by this expression, assuming \( y \) is constant.
Similarly, the partial derivative with respect to \( y \), \( \frac{\partial z}{\partial y} \), gives us \( -x\cdot e^{1-xy} \). With partial derivatives, we can later use these results in the chain rule to find how \( z \) changes over time with respect to another parameter.
In practice, this is like asking: "What happens to \( z \) if only \( x \) changes while \( y \) stays the same?" or vice versa. For our function, we calculated the partial derivative with respect to \( x \), \( \frac{\partial z}{\partial x} \), resulting in \( -y\cdot e^{1-xy} \). This shows that for a small change in \( x \), \( z \) will change by this expression, assuming \( y \) is constant.
Similarly, the partial derivative with respect to \( y \), \( \frac{\partial z}{\partial y} \), gives us \( -x\cdot e^{1-xy} \). With partial derivatives, we can later use these results in the chain rule to find how \( z \) changes over time with respect to another parameter.
Derivative with Respect to a Parameter
The derivative with respect to a parameter is necessary when we are interested in understanding how a dependent variable changes over time or some other changing parameter.
In our example, we needed to find \( \frac{dz}{dt} \), where both \( x \) and \( y \) are functions of \( t \). This requires using the chain rule for differentiation, which helps in dealing with composite functions. The chain rule states:
This relationship uses the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) that we derived earlier, alongside the derivatives \( \frac{dx}{dt} = \frac{1}{3}t^{-2/3} \) and \( \frac{dy}{dt} = 3t^2 \), which reflect the change in \( x \) and \( y \) with \( t \).
By plugging these into the chain rule format, one can compute \( \frac{dz}{dt} \) effectively, thus understanding how \( z \) evolves as \( t \) varies.
In our example, we needed to find \( \frac{dz}{dt} \), where both \( x \) and \( y \) are functions of \( t \). This requires using the chain rule for differentiation, which helps in dealing with composite functions. The chain rule states:
- \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \)
This relationship uses the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) that we derived earlier, alongside the derivatives \( \frac{dx}{dt} = \frac{1}{3}t^{-2/3} \) and \( \frac{dy}{dt} = 3t^2 \), which reflect the change in \( x \) and \( y \) with \( t \).
By plugging these into the chain rule format, one can compute \( \frac{dz}{dt} \) effectively, thus understanding how \( z \) evolves as \( t \) varies.
Exponential Functions
Exponential functions, such as \( e^{1-xy} \) in our equation for \( z \), are powerful mathematical tools that describe situations where growth or decay is proportional to the current amount.
The base of natural exponential functions is \( e \), approximately equal to 2.718. It is unique because its derivative is the same as the function itself, which provides simplicity when deriving expressions that involve \( e \).
Within our problem, the exponential function evaluates the expression \( 1-xy \), which scales the impact of the exponential behavior based on the values of \( x \) and \( y \). In essence, the exponent affects how the function grows or decays based on the combined interaction of \( x \) and \( y \).
The base of natural exponential functions is \( e \), approximately equal to 2.718. It is unique because its derivative is the same as the function itself, which provides simplicity when deriving expressions that involve \( e \).
Within our problem, the exponential function evaluates the expression \( 1-xy \), which scales the impact of the exponential behavior based on the values of \( x \) and \( y \). In essence, the exponent affects how the function grows or decays based on the combined interaction of \( x \) and \( y \).
- For instance, when differentiating, the exponential itself remains, while we take into account the derivative of the exponent (like \(-xy\) in our partial derivatives).