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Use Theorem 15.7 to find the following derivatives. $$d V / d t, \text { where } V=x y z, x=e^{t}, y=2 t+3, \text { and } z=\sin t$$

Short Answer

Expert verified
Answer: \(\frac{dV}{dt} = e^t(2t+3)\sin t + 2e^t\sin t + e^t(2t+3)\cos t\)

Step by step solution

01

Find the partial derivatives of V

First, we will find the partial derivatives of V with respect to x, y, and z. $$\frac{\partial V}{\partial x} = yz$$ $$\frac{\partial V}{\partial y} = xz$$ $$\frac{\partial V}{\partial z} = xy$$
02

Find the derivatives of x, y, and z with respect to t

Now, we will find the derivatives of x, y, and z functions with respect to t: $$\frac{dx}{dt} = \frac{d}{dt}(e^t) = e^t$$ $$\frac{dy}{dt} = \frac{d}{dt}(2t + 3) = 2$$ $$\frac{dz}{dt} = \frac{d}{dt}(\sin t) = \cos t$$
03

Apply the Chain Rule and find \(\frac{dV}{dt}\)

Now, we will substitute the partial derivatives and the derivatives into the Chain Rule expression: $$\frac{dV}{dt} = \frac{\partial V}{\partial x}\frac{dx}{dt} + \frac{\partial V}{\partial y}\frac{dy}{dt} + \frac{\partial V}{\partial z}\frac{dz}{dt}$$ $$\frac{dV}{dt} = (yz)(e^t) + (xz)(2) + (xy)(\cos t)$$ Substitute the given functions for x, y, and z: $$\frac{dV}{dt} = [(2t+3)(\sin t)](e^t) + [(e^t)(\sin t)](2) + [(e^t)(2t+3)](\cos t)$$ Finally, the derivative of V with respect to t is: $$\frac{dV}{dt} = e^t(2t+3)\sin t + 2e^t\sin t + e^t(2t+3)\cos t$$

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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 an essential concept in multivariable calculus, which allow us to examine the rate of change of a function with respect to one variable, while keeping the other variables constant. Imagine you are at a mountain campsite, and you want to know how steep the incline is heading east without worrying about the change in elevation going north. This is what partial derivatives allow us to do but in the mathematical landscape.

In our example, the function V represents the volume, dependent on three other variables—x, y, and z. To find the partial derivative of V with respect to x, denoted as \( \frac{\partial V}{\partial x} \), we treat y and z as constants and only differentiate V with respect to x. The result, for this case, is yz, indicating that when x changes slightly, the volume changes by yz units, assuming y and z stay fixed.

Similarly, \( \frac{\partial V}{\partial y} \) and \( \frac{\partial V}{\partial z} \) tell us how the volume changes with respect to y and z respectively, again treating other variables as constants. To ensure a comprehensive understanding, it's useful for students to practice finding partial derivatives for a variety of functions.
Chain Rule
The chain rule is a powerful tool in calculus, specifically when dealing with functions that are compositions of other functions—essentially 'nested' functions. It reveals how the derivative of the composite relates to the derivatives of each function individually. Think of it as understanding how a system reacts when you change its input, knowing how each component of the system individually responds to changes.

When applying the chain rule in multivariable calculus, we consider how each individual variable's change affects the overall function. In the exercise, we see that V is a function of x, y, and z, which themselves are functions of t. The chain rule tells us that to find \( \frac{dV}{dt} \), we must take each partial derivative of V—found in the first step—and multiply it by the derivative of its corresponding variable with respect to t, then sum these products.

This method is frequently visualized as a tree diagram with branches showing paths of dependence, but the essence is in understanding how changing an 'upstream' variable, like t, affects a 'downstream' outcome, like V through the intermediary variables x, y, and z.
Derivative of Exponential Functions
The derivative of exponential functions reflects constant proportional growth, which makes these functions unique. Exponential functions often appear in real-world applications such as population growth, radioactive decay, and interest calculations. Mathematically, the base e is especially significant owing to its natural properties in calculus.

In our exercise, the function x is given as \( e^t \), which follows the rule that the derivative of \( e^t \) with respect to t is simply \( e^t \) itself. This means for any small change in t, the rate of change of \( e^t \) is equal to its current value. Students may find it useful to remember that the natural exponential function is its own derivative, a unique feature that simplifies many calculus problems. By connecting this property back to real-life exponential growth scenarios, the abstract concept becomes more tangible and easier to grasp.

Understanding the derivative of exponential functions like \( e^t \) is critical as it forms the basis for more complex operations such as solving differential equations and exponential models in various fields of science and engineering.

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Most popular questions from this chapter

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