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Chapter 12: Problem 15

Use Theorem 12.7 to find the following derivatives. When feasible, express your answer in terms of the independent variable. $$d U / d t, \text { where } U=\ln (x+y+z), x=t, y=t^{2}, \text { and } z=t^{3}$$

Short Answer

Expert verified
Answer: The derivative of U(t) with respect to t is dU/dt = (1 + 2t + 3t^2)/(t + t^2 + t^3).

Step by step solution

01

Define the functions

First, define the functions as follows: $$U(t) = \ln(x(t) + y(t) + z(t))$$ where \(x(t) = t\), \(y(t) = t^2\), and \(z(t) = t^3\).
02

Compute the derivatives of x(t), y(t), and z(t) with respect to t

Now compute the derivatives of x(t), y(t), and z(t) with respect to t: $$\frac{dx}{dt} = 1$$ $$\frac{dy}{dt} = 2t$$ $$\frac{dz}{dt} = 3t^2$$
03

Compute the derivative of U(t) with respect to the sum of x(t), y(t), and z(t)

Compute the derivative of the natural logarithm function with respect to the sum of x(t), y(t), and z(t): $$\frac{dU}{d(x+y+z)} = \frac{1}{x+y+z}$$
04

Apply the Chain Rule

Now apply the Chain Rule to find the derivative of U(t) with respect to t: $$\frac{dU}{dt} = \frac{dU}{d(x+y+z)} \cdot \frac{d(x+y+z)}{dt}$$ $$\frac{dU}{dt} = \frac{1}{x+y+z} \cdot (1 + 2t + 3t^2)$$
05

Substitute the expressions for x(t), y(t), and z(t) back into the equation

Substitute the expressions for x(t), y(t), and z(t) back into the equation to express the derivative of U in terms of the independent variable t: $$\frac{dU}{dt} = \frac{1}{t + t^2 + t^3} \cdot (1 + 2t + 3t^2)$$
06

The final answer

The derivative of U with respect to t is: $$\frac{dU}{dt} = \frac{1 + 2t + 3t^2}{t + t^2 + t^3}$$

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Key Concepts

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

Derivative of Natural Logarithm
The derivative of the natural logarithm, denoted as \( \ln(x) \), is a fundamental concept in calculus. To understand this concept, let's take a brief step back and remind ourselves that a logarithm is an exponent to which a base has to be raised to produce a given number. Now, the natural logarithm specifically has the base 'e', where 'e' is an irrational constant approximately equal to 2.71828.

When you differentiate \( \ln(x) \), you are essentially asking how quickly the value of the natural logarithm changes as you slightly change 'x'. It turns out that the rate of this change, or the derivative, is \( \frac{1}{x} \). In other words, the steeper the incline of the natural logarithm at a point, the larger the value of \( \frac{1}{x} \). It鈥檚 important because this derivative appears commonly when dealing with growth processes, including those found in physics, biology, and economics.

By incorporating this understanding, students can better tackle problems involving the natural logarithm in calculus, especially when combined with the chain rule to manage more complex situations where you have \( \ln(f(x)) \), not just \( \ln(x) \).
Implicit Differentiation
Implicit differentiation is a powerful technique used in calculus to find the derivative of an equation where the variables cannot be easily separated. Rather than solving for one variable in terms of another and then differentiating, you differentiate both sides of the equation as they are, with respect to the independent variable.

In essence, you're acknowledging that one variable is a function of another, even if you don't explicitly have the equation for that function. When you come across terms involving the dependent variable, you just tack on a \( \frac{dy}{dx} \) to indicate its derivative with respect to \( x \). This is reflective of the chain rule where, for every function \( u(x) \), its derivative \( \frac{du}{dx} \) automatically accounts for the derivative of \( u \) with respect to \( x \).

Applying implicit differentiation requires a keen eye for functions within functions and a strong grasp of basic differentiation rules. For students, mastering this technique is much like adding a Swiss Army knife to their mathematical toolkit - versatile and applicable in various scenarios, including geometry, physics, and economics.
Calculus Problem Solving
Calculus problem solving encompasses a wide range of techniques and strategies to tackle the wide array of problems you can encounter in the subject. From basic derivative calculations to integration, and from limits to series, approaching calculus problems thoughtfully is key to success.

Some general strategies for calculus problem solving include understanding the problem, identifying what is being asked, choosing the right method for the task (like differentiating or integrating), and simplifying the problem where possible. Fundamental to these strategies is a deep knowledge of calculus concepts and rules, such as the product rule, quotient rule, and chain rule.

Recognizing Patterns and Applying Rules

When solving calculus problems, recognizing patterns is crucial. For instance, seeing the underlying structure of a complicated expression can hint at which rule to apply. The chain rule, for example, is used when you have composite functions, and knowing when to use it comes with practice.

Staying Structured in Your Approach

Maintaining a structured approach by writing down all steps methodically helps avoid errors and provides a clear path back if you hit a dead end. Finally, practicing a variety of problems and reviewing mistakes helps build intuition and problem-solving speed, which is invaluable during tests.

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Most popular questions from this chapter

Batting averages in baseball are defined by \(A=x / y,\) where \(x \geq 0\) is the total number of hits and \(y>0\) is the total number of at bats. Treat \(x\) and \(y\) as positive real numbers and note that \(0 \leq A \leq 1.\) a. Use differentials to estimate the change in the batting average if the number of hits increases from 60 to 62 and the number of at bats increases from 175 to 180 . b. If a batter currently has a batting average of \(A=0.350,\) does the average decrease if the batter fails to get a hit more than it increases if the batter gets a hit? c. Does the answer to part (b) depend on the current batting average? Explain.

Two resistors in an electrical circuit with resistance \(R_{1}\) and \(R_{2}\) wired in parallel with a constant voltage give an effective resistance of \(R,\) where \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\). a. Find \(\frac{\partial R}{\partial R_{1}}\) and \(\frac{\partial R}{\partial R_{2}}\) by solving for \(R\) and differentiating. b. Find \(\frac{\partial R}{\partial R_{1}}\) and \(\frac{\partial R}{\partial R_{2}}\) by differentiating implicitly. c. Describe how an increase in \(R_{1}\) with \(R_{2}\) constant affects \(R\). d. Describe how a decrease in \(R_{2}\) with \(R_{1}\) constant affects \(R\).

Identify and briefly describe the surfaces defined by the following equations. $$-y^{2}-9 z^{2}+x^{2} / 4=1$$

Flow in a cylinder Poiseuille's Law is a fundamental law of fluid dynamics that describes the flow velocity of a viscous incompressible fluid in a cylinder (it is used to model blood flow through veins and arteries). It says that in a cylinder of radius \(R\) and length \(L,\) the velocity of the fluid \(r \leq R\) units from the center-line of the cylinder is \(V=\frac{P}{4 L \nu}\left(R^{2}-r^{2}\right),\) where \(P\) is the difference in the pressure between the ends of the cylinder and \(\nu\) is the viscosity of the fluid (see figure). Assuming that \(P\) and \(\nu\) are constant, the velocity \(V\) along the center line of the cylinder \((r=0)\) is \(V=k R^{2} / L,\) where \(k\) is a constant that we will take to be \(k=1.\) a. Estimate the change in the centerline velocity \((r=0)\) if the radius of the flow cylinder increases from \(R=3 \mathrm{cm}\) to \(R=3.05 \mathrm{cm}\) and the length increases from \(L=50 \mathrm{cm}\) to \(L=50.5 \mathrm{cm}.\) b. Estimate the percent change in the centerline velocity if the radius of the flow cylinder \(R\) decreases by \(1 \%\) and the length \(L\) increases by \(2 \%.\) c. Complete the following sentence: If the radius of the cylinder increases by \(p \%,\) then the length of the cylinder must increase by approximately __________ \(\%\) in order for the velocity to remain constant.

Extending Exercise \(62,\) when three electrical resistors with resistance \(R_{1}>0, R_{2}>0,\) and \(R_{3}>0\) are wired in parallel in a circuit (see figure), the combined resistance \(R,\) measured in ohms \((\Omega),\) is given by \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\) Estimate the change in \(R\) if \(R_{1}\) increases from \(2 \Omega\) to \(2.05 \Omega, R_{2}\) decreases from \(3 \Omega\) to \(2.95 \Omega,\) and \(R_{3}\) increases from \(1.5 \Omega\) to \(1.55 \Omega.\)
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