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Chapter 10: Problem 1

Let \(f(x, y)=x^{2}+y^{2}\) with \(x(t)=3 t\) and \(y(t)=e^{t} .\) Find the derivative of \(w=f(x, y)\) with respect to \(t\) when \(t=\ln 2\).

Short Answer

Expert verified
The derivative is \(18(\ln 2) + 8\) when evaluated at \(t = \ln 2\).

Step by step solution

01

Define the Composite Function

The function \(w\) is defined as a composite function: \(w(t) = f(x(t), y(t))\), where \(f(x, y)=x^2 + y^2\), \(x(t)=3t\), and \(y(t)=e^t\). Substitute the expressions for \(x(t)\) and \(y(t)\) into \(f(x, y)\) to get \(w(t) = (3t)^2 + (e^t)^2\). Simplify to find \(w(t) = 9t^2 + e^{2t}\).
02

Differentiate with Respect to t

Take the derivative of \(w(t) = 9t^2 + e^{2t}\) with respect to \(t\). The derivative \(\frac{dw}{dt} = \frac{d}{dt}(9t^2) + \frac{d}{dt}(e^{2t})\). Use the power rule for \(9t^2\) to get \(18t\). For \(e^{2t}\), use the chain rule: the derivative is \(2e^{2t}\). Combining these, \(\frac{dw}{dt} = 18t + 2e^{2t}\).
03

Evaluate the Derivative at t=ln(2)

Substitute \(t = \ln 2\) into the derivative equation: \(\frac{dw}{dt} = 18(\ln 2) + 2e^{2(\ln 2)}\). Simplify \(2e^{2(\ln 2)}\) using the property \(e^{2(\ln 2)} = (e^{\ln 2})^2 = 2^2 = 4\). Therefore, \(\frac{dw}{dt} = 18(\ln 2) + 8\).
04

Simplify the Expression

The expression to evaluate is \(18(\ln 2) + 8\). Since \(\ln 2\) is a constant, this is the final form of the derivative. Calculate numerically if needed, recognizing that \(\ln 2\) is approximately 0.693.

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Key Concepts

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

Composite Function
In the realm of multivariable calculus, a composite function is essentially a function within a function. Imagine you're baking a complex cake. Each layer is made from specific ingredients, but they all come together to form the final delicious dessert. Similarly, here is the concept of a composite function.
  • A composite function involves an outer function and an inner function, such that the output of the inner function becomes the input to the outer function.
  • In the problem given, the composite function is defined as \( w(t) = f(x(t), y(t)) \).
  • This means that the variables \( x(t) \) and \( y(t) \) are plugged into another function \( f(x, y) \) to generate a new function depending only on \( t \).
In simple terms, it's like using a shipping label to create a personalized package. Each representing a line of processors where materials are refined into a polished product that goes out into the world.
Chain Rule
The chain rule is a pivotal technique in calculus used when differentiating composite functions. It is like a magician's secret trick for unveiling the derivative of nestled functions.
  • The chain rule helps to compute the derivative of a composition by multiplying the derivative of the outer function by the derivative of the inner function.
  • In our exercise, when dealing with \( e^{2t} \), the chain rule comes into play because the exponent itself is a function of \( t \).
  • This means the derivative is calculated by taking the derivative of \( e^{u} \) with respect to \( u \) (which is \( e^{u} \)), and then multiplying by the derivative of \( u \) with respect to \( t \), where \( u = 2t \).
  • As a result, the derivative becomes \( 2e^{2t} \).
The chain rule is your best friend in calculus-manipulating situations where functions are layered like onions, helping you peel back each layer smoothly.
Differentiation
Differentiation is a fundamental concept in calculus that is used to compute the derivative of a function. This is akin to measuring how a shadow lengthens and shortens as the sun moves across the sky.
  • The derivative represents the rate at which a variable quantity changes with respect to another variable.
  • In our example, the function \( w(t) = 9t^2 + e^{2t} \) is differentiated with respect to \( t \). The goal is to find how \( w \) changes as \( t \) changes.
  • Different rules apply during differentiation, such as the power rule for \( 9t^2 \), where the exponent is brought in front and the power itself is reduced by one, resulting in \( 18t \).
  • Similarly, exponential functions like \( e^{2t} \) require careful application of the chain rule as previously described.
Differentiation is key to uncovering intricate relationships between variables. It enables you to measure subtle shifts, much like tuning in to subtle whispers in the grand conversation of mathematics.

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Most popular questions from this chapter

Refer to the negative binomial host-parasitoid model. Problems \(7,8,11\), and 12 are best done with the help of a spreadsheet, but can also be done with a calculator. The negative binomial model is a discrete-generation host- parasitoid model of the form $$ \begin{array}{l} N_{t+1}=b N_{t}\left(1+\frac{a P_{t}}{k}\right)^{-k} \\ P_{t+1}=c N_{t}\left[1-\left(1+\frac{a P_{t}}{k}\right)^{ k}\right] \end{array} $$ for \(t=0,1,2, \ldots\) Show that when the initial parasitoid density is \(P_{0}=0\), the negative binomial model reduces to $$ N_{t+1}=b N_{t} $$ With \(N_{0}\) denoting the initial host density, find an expression for \(N_{t}\) in terms of \(N_{0}\) and the parameter \(b\).

The functional responses of some predators are sigmoidal; that is, the number of prey attacked per predator as a function of prey density has a sigmoidal shape. If we denote the prey density by \(N\), the predator density by \(P\), the time available for searching for prey by \(T\), and the handling time of each prey item per predator by \(T_{h}\), then the number of prey encounters per predator as a function of \(N, T\), and \(T_{h}\) can be expressed as $$ f\left(N, T, T_{h}\right)=\frac{b^{2} N^{2} T}{1+c N+b T_{h} N^{2}} $$ where \(b\) and \(c\) are positive constants. (a) Investigate how an increase in the prey density \(N\) affects the function \(f\left(N, T, T_{h}\right)\) (b) Investigate how an increase in the time \(T\) available for search affects the function \(f\left(N, T, T_{h}\right)\). (c) Investigate how an increase in the handling time \(T_{h}\) affects the function \(f\left(N, T, T_{h}\right)\) (d) Graph \(f\left(N, T, T_{h}\right)\) as a function of \(N\) when \(T=2.4\) hours, \(T_{h}=0.2\) hours, \(b=0.8\), and \(c=0.5\)

In what direction does \(f(x, y)=e^{x} \cos y\) increase most rapidly at \((0, \pi / 2) ?\)

In the Nicholson-Bailey model, the fraction of hosts escaping parasitism is given by $$ f(P)=e^{-a P} $$ (a) Graph \(f(P)\) as a function of \(P\) for \(a=0.1\) and \(a=0.01\). (b) For a given value of \(P\), how are the chances of escaping parasitism affected by increasing \(a\) ?

Find the gradient of each function. $$ f(x, y)=\tan \frac{x-y}{x+y} $$
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