Problem 14 - Calculus Early Transcendentals

Question

Let \(W(s, t)=F(u(s, t), v(s, t)),\) where \(F, u,\) and \(v\) are differentiable, and $$\begin{array}{ll}{u(1,0)=2} & {v(1,0)=3} \\ {u_{s}(1,0)=-2} & {v_{s}(1,0)=5} \\ {u_{t}(1,0)=6} & {v_{t}(1,0)=4} \\ {F_{v}(2,3)=-1} & {F_{r}(2,3)=10}\end{array}$$ Find \(W_{s}(1,0)\) and \(W_{t}(1,0)\)

Step-by-Step Solution

Answer

\( W_{s}(1,0) = -25 \) and \( W_{t}(1,0) = 56 \).

Step 1: Understand the Problem

We have a composite function \( W(s, t) = F(u(s, t), v(s, t)) \), and we need to find its partial derivatives with respect to \( s \) and \( t \) at the point \((1, 0)\). We have the derivatives of \( u \) and \( v \) with respect to \( s \) and \( t \), and we need to use the chain rule for partial derivatives.

Step 2: Recall the Chain Rule

To find \( W_{s}(s,t) \), apply the chain rule for multi-variable functions: \[ W_{s}(s,t) = F_{u}(u,v)\frac{\partial u}{\partial s} + F_{v}(u,v)\frac{\partial v}{\partial s} \]. Similarly, \( W_{t}(s,t) = F_{u}(u,v)\frac{\partial u}{\partial t} + F_{v}(u,v)\frac{\partial v}{\partial t} \).

Step 3: Substitute Known Values for \( W_{s}(1,0) \)

Using the given values: \( u(1,0) = 2 \), \( v(1,0) = 3 \), \( u_{s}(1,0) = -2 \), \( v_{s}(1,0) = 5 \), \( F_{u}(2,3) = 10 \), and \( F_{v}(2,3) = -1 \), we substitute them into the formula:\[W_{s}(1,0) = 10 \cdot (-2) + (-1) \cdot 5 = -20 - 5 = -25.\]

Step 4: Substitute Known Values for \( W_{t}(1,0) \)

Using the given values: \( u_{t}(1,0) = 6 \) and \( v_{t}(1,0) = 4 \), substitute them along with \( F_{u}(2,3) \) and \( F_{v}(2,3) \) into the formula:\[W_{t}(1,0) = 10 \cdot 6 + (-1) \cdot 4 = 60 - 4 = 56.\]

Step 5: Conclusion

After evaluating the calculations for the partial derivatives: \( W_{s}(1,0) = -25 \) and \( W_{t}(1,0) = 56 \).

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used to differentiate composite functions. In multivariable calculus, the rule extends to functions with more than one variable. This is useful when computing the derivative of a function that depends on other functions. Consider a composite function like \( W(s, t) = F(u(s, t), v(s, t)) \). Here, \( W \) is a function of \( s \) and \( t \), but it depends on \( u \) and \( v \), which are themselves functions of \( s \) and \( t \).

• The Chain Rule allows us to find how changes in \( s \) or \( t \) affect \( W \) through their relationships with \( u \) and \( v \).
• To apply the Chain Rule in this context, we use partial derivatives: \( W_s(s,t) \) and \( W_t(s,t) \).
• For example, \( W_s(s,t) = F_{u}(u,v) \cdot \frac{\partial u}{\partial s} + F_{v}(u,v) \cdot \frac{\partial v}{\partial s} \).

Breaking the problem into smaller derivatives helps track how functions interconnect and change relative to each variable.

Composite Function

A composite function is a combination of two or more functions where the output of one function becomes the input of another. In our example, \( W(s, t) = F(u(s, t), v(s, t)) \), \( F \) is a function of \( u \) and \( v \), while \( u \) and \( v \) themselves are functions of \( s \) and \( t \).

• Composite functions are often encountered in problems where outputs from one stage or step are inputs for another. This can make calculations more complex, hence the importance of the Chain Rule to unravel the dependencies.
• Understanding the nature of each component function and its domain is crucial in correctly using composite functions.
• In multivariable calculus, composite functions allow us to work with structures that model real-world phenomena, where multiple factors in tandem influence an outcome.

Proper handling of composite functions involves careful organization of each function's derivatives, ensuring each link in the chain is correctly evaluated.

Partial Derivatives

Partial derivatives are used to take the derivative of functions with more than one variable concerning one variable at a time, treating the others as constants. In our case, we focused on calculating \( W_s \) and \( W_t \), which are partial derivatives of the composite function \( W(s, t) \).

• \( W_s \) is the rate at which \( W \) changes with respect to \( s \), holding \( t \) constant.
• \( W_t \) indicates how \( W \) varies as \( t \) changes, while \( s \) is constant.
• This technique is invaluable in multivariable calculus, giving insights into how each variable contributes independently to the function’s change.

Challenges arise in ensuring all necessary derivatives are computed properly before substitution into a formula. It's essential to plug in correct values, particularly in a detailed problem like finding \( W_s(1,0) \) and \( W_t(1,0) \), for precise solutions.