Problem 8
Question
In Exercises \(7-12,\) functions \(z=f(x, y), x=g(t)\) and \(y=h(t)\) are given. (a) Use the Multivariable Chain Rule to compute \(\frac{d z}{d t}\). (b) Evaluate \(\frac{d z}{d t}\) at the indicated \(t\) -value. $$ z=x^{2}-y^{2}, \quad x=t, \quad y=t^{2}-1 ; \quad t=1 $$
Step-by-Step Solution
Verified Answer
The value of \(\frac{d z}{d t}\) at \(t = 1\) is 2.
1Step 1: Determine the Partial Derivatives
To find \(\frac{d z}{d t}\), we apply the multivariable chain rule. Start by calculating the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). For \(z=x^2-y^2\), we have: \[\frac{\partial z}{\partial x} = 2x\] \[\frac{\partial z}{\partial y} = -2y\]
2Step 2: Express Partial Derivatives in Terms of t
Since \(x = g(t) = t\) and \(y = h(t) = t^2 - 1\), substitute these into the partial derivatives. Thus, \(\frac{\partial z}{\partial x} = 2t\) and \(\frac{\partial z}{\partial y} = -2(t^2 - 1)\).
3Step 3: Calculate Derivatives of x and y with Respect to t
Find \(\frac{d x}{d t}\) and \(\frac{d y}{d t}\). Given \(x = t\), we have: \[\frac{d x}{d t} = 1\] And for \(y = t^2 - 1\), we have: \[\frac{d y}{d t} = 2t\]
4Step 4: Apply the Multivariable Chain Rule
The chain rule for \(\frac{d z}{d t}\) gives: \[\frac{d z}{d t} = \frac{\partial z}{\partial x}\frac{d x}{d t} + \frac{\partial z}{\partial y}\frac{d y}{d t}\] Plug in the values: \[= 2t \cdot 1 + (-2(t^2 - 1)) \cdot 2t\] \[= 2t - 4t(t^2 - 1)\]
5Step 5: Simplify the Expression
Simplify \(\frac{d z}{d t} = 2t - 4t(t^2 - 1)\): \[= 2t - 4t^3 + 4t\] \[= 6t - 4t^3\] Factor: \[= 2t(3 - 2t^2)\]
6Step 6: Evaluate at t = 1
To find \(\frac{d z}{d t}\) at \(t = 1\), substitute 1 into the simplified expression: \[\frac{d z}{d t} \bigg|_{t=1} = 2(1)(3 - 2(1)^2)\] \[= 2(1)(3 - 2) = 2\]
7Step 7: Final Result
The derivative \(\frac{d z}{d t}\) evaluated at \(t = 1\) is 2.
Key Concepts
Partial DerivativesDerivative EvaluationParametric Functions
Partial Derivatives
Partial derivatives are a crucial concept when dealing with multivariable functions. They represent how a function changes as one of its input variables changes, while keeping the other variables constant.
To find the partial derivative with respect to a specific variable, treat all other variables as constants and then differentiate as usual.
For the function \(z = x^2 - y^2\), the partial derivative with respect to \(x\) is given by \(\frac{\partial z}{\partial x} = 2x\). This means that if you slightly increase \(x\) by a small amount, while holding \(y\) constant, \(z\) will increase by approximately \(2x\) times that small amount.
Similarly, the partial derivative with respect to \(y\) is \(\frac{\partial z}{\partial y} = -2y\). This negative sign indicates that if \(y\) increases, \(z\) will decrease, assuming \(x\) is constant.
To find the partial derivative with respect to a specific variable, treat all other variables as constants and then differentiate as usual.
For the function \(z = x^2 - y^2\), the partial derivative with respect to \(x\) is given by \(\frac{\partial z}{\partial x} = 2x\). This means that if you slightly increase \(x\) by a small amount, while holding \(y\) constant, \(z\) will increase by approximately \(2x\) times that small amount.
Similarly, the partial derivative with respect to \(y\) is \(\frac{\partial z}{\partial y} = -2y\). This negative sign indicates that if \(y\) increases, \(z\) will decrease, assuming \(x\) is constant.
Derivative Evaluation
Derivative evaluation is the process of finding the value of a derivative at a specific point. This step helps us understand the instantaneous rate of change of a function at a particular value.
In this exercise, after using the chain rule to find the derivative of \(z\) with respect to \(t\), we obtain an expression \(\frac{d z}{d t} = 6t - 4t^3\).
To evaluate this derivative at \(t = 1\), substitute \(t = 1\) into the expression:
In this exercise, after using the chain rule to find the derivative of \(z\) with respect to \(t\), we obtain an expression \(\frac{d z}{d t} = 6t - 4t^3\).
To evaluate this derivative at \(t = 1\), substitute \(t = 1\) into the expression:
- \(6(1) - 4(1)^3 = 6 - 4 = 2\)
Parametric Functions
Parametric functions involve expressing one or more variables as functions of one or more parameters. This approach is particularly useful when analyzing the behavior of systems described by several interconnected variables.
In the given problem, \(x\) and \(y\) are expressed as functions of \(t\). Specifically, \(x = t\) and \(y = t^2 - 1\).
This setup allows us to track how \(x\) and \(y\) interact and affect \(z\) over time as \(t\) changes. By using parametric expressions, it becomes easier to apply the multivariable chain rule, as it helps bridge the dependence between \(z\) and \(t\) through \(x\) and \(y\).
Parametric functions are prevalent in physics and engineering when describing motion, trajectories, and dynamic systems.
In the given problem, \(x\) and \(y\) are expressed as functions of \(t\). Specifically, \(x = t\) and \(y = t^2 - 1\).
This setup allows us to track how \(x\) and \(y\) interact and affect \(z\) over time as \(t\) changes. By using parametric expressions, it becomes easier to apply the multivariable chain rule, as it helps bridge the dependence between \(z\) and \(t\) through \(x\) and \(y\).
Parametric functions are prevalent in physics and engineering when describing motion, trajectories, and dynamic systems.
Other exercises in this chapter
Problem 7
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