Problem 4
Question
If \(z=f(x, y),\) where \(x=g(t)\) and \(y=h(t),\) we can substitute and write \(z\) as an explicit function of \(t\). T/F: Using the Multivariable Chain Rule to find \(\frac{d z}{d t}\) is sometimes easier than first substituting and then taking the derivative.
Step-by-Step Solution
Verified Answer
True; using the Multivariable Chain Rule is sometimes easier.
1Step 1: Understand the Problem
We need to find the derivative \( \frac{dz}{dt} \) given that \( z = f(x, y) \), with \( x = g(t) \) and \( y = h(t) \). This means \( z \) is implicitly a function of \( t \) through \( x \) and \( y \).
2Step 2: Analyze the Methods
We have two methods: combining \( x = g(t) \) and \( y = h(t) \) to express \( z \) explicitly in terms of \( t \) and then differentiating, or using the multivariable chain rule directly to find \( \frac{dz}{dt} \).
3Step 3: Describe the Multivariable Chain Rule Method
The multivariable chain rule states that \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \). This method takes advantage of partial derivatives of \( f(x, y) \).
4Step 4: Consider Efficiencies
When \( f(x, y) \) is complex or involves intricate functions, using the chain rule can be more straightforward than expressing \( z \) explicitly in terms of \( t \). This avoids complicated substitutions and simplifies computations.
5Step 5: Apply to Example
If \( f(x, y) = x^2 + y^2 \), \( x = t^2 \), \( y = e^t \), then expressing \( z \) in terms of \( t \) could become cumbersome, whereas the chain rule method directly gives \( \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \).
6Step 6: Conclusion
Using the chain rule can be easier, especially when the expressions for \( x(t) \), \( y(t) \), or the function \( f(x, y) \) are complicated, validating the statement presented as true.
Key Concepts
Chain RuleImplicit DifferentiationPartial Derivatives
Chain Rule
In multivariable calculus, the chain rule is a powerful tool used to calculate derivatives of composite functions. When you have a function like \( z = f(x, y) \) and both \( x \) and \( y \) are themselves functions dependent on another variable, say \( t \), it's possible to compute the derivative \( \frac{dz}{dt} \) directly using the chain rule. This bypasses the need to first solve \( z \) explicitly as a function of \( t \), which can become complex and cumbersome.
The formula for the multivariable chain rule is:
Using the chain rule is often an efficient strategy, especially when the given functions involve components that are difficult to simplify or express neatly in terms of \( t \). Applying the chain rule can save time and reduce errors, primarily when managing complex expressions.
The formula for the multivariable chain rule is:
- \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \)
Using the chain rule is often an efficient strategy, especially when the given functions involve components that are difficult to simplify or express neatly in terms of \( t \). Applying the chain rule can save time and reduce errors, primarily when managing complex expressions.
Implicit Differentiation
Implicit differentiation is a technique used when you have an equation in which the dependent variable isn't isolated on one side — essentially, it's when the variable you want to differentiate isn't clearly expressed in terms of another. In the context of functions like \( z = f(x, y) \), where \( x \) and \( y \) are functions of \( t \), implicit differentiation ensures you can still find derivatives despite the lack of an explicit form.
When employing implicit differentiation, you systematically differentiate both sides of an equation concerning the independent variable — here, \( t \).
When employing implicit differentiation, you systematically differentiate both sides of an equation concerning the independent variable — here, \( t \).
- This involves recognizing and performing the chain rule to account for the implicit dependency of \( z \) on \( t \) through \( x \) and \( y \).
Partial Derivatives
Partial derivatives are integral to understanding how functions change as their variables change. When dealing with a multivariable function like \( f(x, y) \), a partial derivative of \( f \) concerning one variable is taken while holding the other constant. This concept plays a crucial role when applying the multivariable chain rule.
For the given function \( z = f(x, y) \):
Understanding partial derivatives help you to analyze and break down changes in complex systems, such as those found in multivariable calculus, aiding the overall calculus toolkit. They facilitate smoother computations and an improved grasp of multidimensional changes.
For the given function \( z = f(x, y) \):
- The partial derivative \( \frac{\partial z}{\partial x} \) reflects the rate of change of \( z \) as \( x \) changes adjacent to the effect of \( y \) held constant.
- Similarly, \( \frac{\partial z}{\partial y} \) examines how \( z \) varies with changes in \( y \), keeping \( x \) constant.
Understanding partial derivatives help you to analyze and break down changes in complex systems, such as those found in multivariable calculus, aiding the overall calculus toolkit. They facilitate smoother computations and an improved grasp of multidimensional changes.
Other exercises in this chapter
Problem 3
\(\mathrm{T} / \mathrm{F}:\) If \(z=f(x, y)\) is differentiable, then the change in \(z\) over small changes \(d x\) and \(d y\) in \(x\) and \(y\) is approxima
View solution Problem 4
Explain what it means to "solve a constrained optimization" problem.
View solution Problem 4
Explain in your own words why we do not refer to the tangent line to a surface at a point, but rather to directional tangent lines to a surface at a point.
View solution Problem 4
Give an example of a closed, unbounded set.
View solution