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Let \(W(s, t)=F(u(s, t), v(s, t))\) where $$\begin{array}{r}u(1,0)=-4, u_{s}(1,0)=-7, u_{t}(1,0)=-4 \\ v(1,0)=8, v_{s}(1,0)=-8, v_{t}(1,0)=3 \\ F_{u}(-4,8)=1, F_{v(-4,8)=5\end{array}$$ \(W_{s}(1,0)= _______ W_{t}(1,0)=\) ____________.

Short Answer

Expert verified
\(W_s(1,0)= -47\) \(W_t(1,0)= 11\)

Step by step solution

01

Identify the given values at the point (1,0)

We are given the following information: \(u(1,0)=-4\), \(u_{s}(1,0)=-7\), \(u_{t}(1,0)=-4\) \(v(1,0)=8\), \(v_{s}(1,0)=-8\), \(v_{t}(1,0)=3\) \(F_{u}(-4,8)=1\), \(F_{v}(-4,8)=5\)
02

Calculate W_s(1,0) using the chain rule

Using the chain rule \(W_s(s,t) = F_u(u(s, t), v(s, t)) \times u_s(s, t) + F_v(u(s, t), v(s, t)) \times v_s(s, t)\) We plug our given values into the equation: \(W_s(1,0) = F_u(-4,8) \times u_s(1,0) + F_v(-4,8) \times v_s(1,0)\) \(W_s(1,0) = (1) \times (-7) + (5) \times (-8)\) \(W_s(1,0) = -7 - 40 = -47\) So, the value of \(W_s(1,0) = -47\).
03

Calculate W_t(1,0) using the chain rule

Using the chain rule \(W_t(s,t) = F_u(u(s, t), v(s, t)) \times u_t(s, t) + F_v(u(s, t), v(s, t)) \times v_t(s, t)\) We plug our given values into the equation: \(W_t(1,0) = F_u(-4,8) \times u_t(1,0) + F_v(-4,8) \times v_t(1,0)\) \(W_t(1,0) = (1) \times (-4) + (5) \times (3)\) \(W_t(1,0) = -4 + 15 = 11\) So, the value of \(W_t(1,0) = 11\).
04

Write down the final answer

We found the values of the partial derivatives of \(W(s, t)\) with respect to s and t at the point (1,0): \(W_s(1,0) = -47\) \(W_t(1,0) = 11\) Thus, the final answer is: \(W_s(1,0)= -47\) \(W_t(1,0)= 11\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
In multivariable calculus, the chain rule is an important technique for finding the derivative of a composite function. It essentially allows you to understand how changes in one set of variables influence another when they are interdependent. Consider a scenario where you have a function of multiple variables, like in our case with the function \( W(s, t) = F(u(s, t), v(s, t)) \).

The chain rule formula for partial derivatives is used to calculate the changes in \( W \) with respect to \( s \) or \( t \) by expressing it as a sum of changes through the dependent variables \( u \) and \( v \):
  • For \( W_s \), we use: \[ W_s(s, t) = F_u(u(s, t), v(s, t)) \times u_s(s, t) + F_v(u(s, t), v(s, t)) \times v_s(s, t) \]
  • For \( W_t \), the formula is: \[ W_t(s, t) = F_u(u(s, t), v(s, t)) \times u_t(s, t) + F_v(u(s, t), v(s, t)) \times v_t(s, t) \]
These expressions provide a way to differentiate composite functions, where each internal function contributes to the rate of change of the outer function.

Understanding how to apply the chain rule helps solve more complex differentiation problems in different contexts.
Partial Derivatives
Partial derivatives are derivatives of functions with multiple variables where we take the derivative with respect to one variable while keeping others constant. This is particularly useful in scenarios such as optimization or analyzing rates of change.

In the context of function \( W(s, t) = F(u(s, t), v(s, t)) \), we have partial derivatives such as \( u_s \), \( u_t \), \( v_s \), and \( v_t \). Each represents the rate of change of the respective function with respect to \( s \) or \( t \):
  • \( u_s \) is the partial derivative of \( u \) with respect to \( s \).
  • \( u_t \) is the partial derivative of \( u \) with respect to \( t \).
  • \( v_s \) is the partial derivative of \( v \) with respect to \( s \).
  • \( v_t \) is the partial derivative of \( v \) with respect to \( t \).
Each of these derivatives measures how a small change in one input variable changes the function value. Partial derivatives are foundational in understanding how complex functions behave when subjected to different input changes, which is essential in fields like engineering and physics.

By evaluating these at specific points, we obtain insights into the function's behavior at those points, crucial for applying the chain rule effectively.
Composite Functions
Composite functions, or functions of a function, are a set of deeply nested functions where one function is input for another. In multivariable calculus, this nesting can involve multiple layers and variables. The function \( W(s, t) = F(u(s, t), v(s, t)) \) is a prime example of a composite function.

Here, we have:
  • \( u(s, t) \) and \( v(s, t) \) are the inner functions, each dependent on variables \( s \) and \( t \) from an outer scope.
  • \( F(u, v) \) is the outer function that relies on the outputs of \( u \) and \( v \) as its inputs.
The key task with composite functions is to understand how changes in the innermost variables ripple through to affect the outer result.

This is where the chain rule shines, by helping calculate derivatives of these layered structures, guiding us in unraveling the complex dependency web. By plugging specific values of the smaller components, you get the behavior of the entire system at particular points.

Understanding composite functions is vital for calculating derivatives in practical applications like modeling natural processes or optimizing systems in mathematics and science.

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Most popular questions from this chapter

A rectangular box with length \(x,\) width \(y,\) and height \(z\) is being built. The box is positioned so that one corner is stationed at the origin and the box lies in the first octant where \(x, y,\) and \(z\) are all positive. There is an added constraint on how the box is constructed: it must fit underneath the plane with equation \(x+2 y+3 z=6\). In fact, we will assume that the corner of the box "opposite" the origin must actually lie on this plane. The basic problem is to find the maximum volume of the box. a. Sketch the plane \(x+2 y+3 z=6,\) as well as a picture of a potential box. Label everything appropriately. b. Explain how you can use the fact that one corner of the box lies on the plane to write the volume of the box as a function of \(x\) and \(y\) only. Do so, and clearly show the formula you find for \(V(x, y)\). c. Find all critical points of \(V\). (Note that when finding the critical points, it is essential that you factor first to make the algebra easier.) d. Without considering the current applied nature of the function \(V\), classify each critical point you found above as a local maximum, local minimum, or saddle point of \(V\). e. Determine the maximum volume of the box, justifying your answer completely with an appropriate discussion of the critical points of the function. f. Now suppose that we instead stipulated that, while the vertex of the box opposite the origin still had to lie on the plane, we were only going to permit the sides of the box, \(x\) and \(y,\) to have values in a specified range (given below). That is, we now want to find the maximum value of \(V\) on the closed, bounded region $$\frac{1}{2} \leq x \leq 1, \quad 1 \leq y \leq 2$$ Find the maximum volume of the box under this condition, justifying your answer fully.

Let \(z=f(x, y)=50-(x+1)^{2}-(y+3)^{2}\) and \(z=h(x, y)=24-2 x-6 y\). Suppose a person is walking on the surface \(z=f(x, y)\) in such a way that she walks the curve which is the intersection of \(f\) and \(h\). a. Show that \(x(t)=4 \cos (t)\) and \(y(t)=4 \sin (t)\) is a parameterization of the "shadow" in the \(x y\) -plane of the curve that is the intersection of the graphs of \(f\) and \(h\). b. Use the parameterization from part (a) to find the instantaneous rate at which her height is changing with respect to time at the instant \(t=2 \pi / 3\).

Use Lagrange multipliers to find the maximum and minimum values of \(f(x, y)=3 x-4 y\) subject to the constraint \(x^{2}+3 y^{2}=129,\) if such values exist. maximum \(=\) __________. minimum \(=\)___________.

Given \(F(r, s, t)=r\left(9 s^{4}-t^{5}\right),\) compute: $$F_{r s t}=$$ _________.

Let \(z=g(u, v)\) and \(u(r, s), v(r, s)\). How many terms are there in the expression for \(\partial z / \partial r ?\) __________ terms

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