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In Exercises 23 and \(24,\) a particle is moving along the curve \(y=f(x) .\) \(y=f(x)=\frac{10}{1+x^{2}}\) If \( \)d x / d t=3 \mathrm{cm} / \mathrm{sec}, \text { find } d y / d t \(d x / d t=3 \mathrm{cm} / \mathrm{sec},\) find \(d y / d t\) at the point where $$x=-2 \text { . } \quad \text { (b) } x=0 . \quad \text { (c) } x=20$$

Short Answer

Expert verified
The rate of change of \(y\) with respect to \(t\) , \(dy/dt\) at where \(x=-2\) is \(4.8 cm/sec\), where \(x=0\) is \(0 cm/sec\), and where \(x=20\) is \(-0.03 cm/sec\).

Step by step solution

01

Differentiate the Given Function

First, the given function \(y=f(x)=\frac{10}{1+x^{2}}\) should be differentiated with respect to \(x\). The derivative will be given by \(f'(x) = \frac{d}{dx} [\frac{10}{1+x^2}]\). By applying the quotient rule, which states that the derivative of \(\frac{u}{v}\) is \(\frac{vu'-uv'}{v^2}\). Here \(u=10\) and \(v=1+x^2\). Hence, \(f'(x) = \frac{(1+x^2)(0) - 10(2x)}{(1+x^2)^2} = -\frac{20x}{(1+x^2)^2}\)
02

Apply the Chain Rule

By applying the chain rule, which states that for a composition of functions, the derivative of outside function times the derivative of inside function, \(dy/dt = f'(x) \cdot dx/dt\)
03

Substitute the Values for \(x = -2, 0, 20\)

Now, we substitute the values of \(x\) into \(dy/dt\) formula and find \(dy/dt\) for each. For \(x = -2\), \(dy/dt = \frac{-20 \cdot -2}{(1+(-2)^2)^2} \cdot 3 = 3 \cdot \frac{40}{25} = 4.8\). Similarly, for \(x = 0\), \(dy/dt = \frac{-20 \cdot 0}{(1+0^2)^2} \cdot 3 = 0\). Lastly for \(x = 20\), \(dy/dt = \frac{-20 \cdot 20}{(1+20^2)^2} \cdot 3 = -0.03\). Hence, at the point where \(x=-2, dy/dt=4.8cm/sec\), at the point where \(x=0, dy/dt=0cm/sec\), and lastly at the point where \(x=20, dy/dt=-0.03cm/sec\).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus that deals with how a function changes as its input changes. It's like finding the rate at which one quantity changes with respect to another. In this exercise, we started with the function \(y = f(x) = \frac{10}{1+x^2}\). Our task was to find how \(y\) changes as \(x\) changes, which means calculating \(\frac{dy}{dx}\). This is accomplished using differentiation. By knowing \(\frac{dx}{dt} = 3\, \text{cm/sec}\) (the rate at which \(x\) changes with time), we can further find out how \(y\) changes over time (\(\frac{dy}{dt}\)). Differentiation helps us in understanding these dynamic changes by providing the slope of the function at any given point.
Understanding differentiation is crucial because it forms the basis for other complex calculus concepts. Remember, it's all about understanding how tiny changes in input create changes in output!
Chain Rule
The Chain Rule is a method in calculus used to differentiate composite functions. Whenever you have a function within another function, the Chain Rule tells us how to find the derivative. In simpler terms, it's about peeling away layers like an onion to find the rate of change of the whole structure.
In our problem, after finding \(f'(x)\), we applied the Chain Rule to express \(\frac{dy}{dt}\). The Chain Rule formula states: for a function \(y = g(f(x))\), \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\). Applying this rule helps us connect how \(y\) changes over time through its dependence on \(x\). It's a valuable tool, as seen, allowing us to translate changes in one variable to another.
  • Use the derivative of the outer function.
  • Multiply by the derivative of the inner function.
Mastering the Chain Rule opens up a wider understanding of relationships between different variables in dynamic systems.
Quotient Rule
The Quotient Rule is essential when differentiating functions divided by each other. In our case, we have \(y = \frac{10}{1+x^2}\), a quotient of two simpler functions. The rule provides a formula to find the derivative without separately differentiating the numerator and the denominator. For differentiation of \(\frac{u}{v}\), the rule is:
  • First, let \(u = 10\) and \(v = 1+x^2\).
  • Find the derivatives \(u'\) and \(v'\).
  • Apply the formula: \(\frac{dy}{dx} = \frac{v \cdot u'- u \cdot v'}{v^2}\).

Through this method, we derive \(-\frac{20x}{(1+x^2)^2}\), capturing how \(y\) changes with respect to \(x\). The Quotient Rule is powerful because many real-world problems involve ratios and fractions, making differentiation clear and straightforward for these cases.
Particle Motion
Particle motion in calculus involves understanding how the position of a particle changes over time, which is essential in physics and engineering. In our scenario, the particle moves along the curve defined by \(y=f(x)\), and we need to find how quickly its vertical position (\(y\)) changes as its horizontal position (\(x\)) changes.
Given that \(\frac{dx}{dt} = 3\, \text{cm/sec}\), we tracked how \(y\) changes with time by calculating \(\frac{dy}{dt}\) at various points, like \(x = -2, 0, \) and \(20\). Understanding particle motion requires knowing both the position and how it evolves.
  • Identify your key positions: here \(x\).
  • Apply related rates tactics using derived formulas.
By discerning the rate of change at these points, we could understand how the particle's motion alters—useful for predicting motion in dynamic systems.

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Most popular questions from this chapter

Multiples of \(P i\) Store any number as \(X\) in your calculator. Then enter the command \(X-\tan (X) \rightarrow X\) and press the ENTER key repeatedly until the displayed value stops changing. The result is always an integral multiple of \(\pi .\) Why is this so? [Hint: These are zeros of the sine function.]

Let \(y=f(x)=x^{3}-4 x\) If \(d x / d t=-2 \mathrm{cm} / \mathrm{sec},\) find \(d y / d t\) at the point where (a) \(x=-3 . \quad\) (b) \(x=1 . \quad\) (c) \(x=4\)

Unique Solution Assume that \(f\) is continuous on \([a, b]\) and differentiable on \((a, b) .\) Also assume that \(f(a)\) and \(f(b)\) have op- posite signs and \(f^{\prime} \neq 0\) between \(a\) and \(b\) . Show that \(f(x)=0\) exactly once between \(a\) and \(b .\)

You may use a graphing calculator to solve the following problems. True or False Newton's method will not find the zero of \(f(x)=x /\left(x^{2}+1\right)\) if the first guess is greater than \(1 .\) Justify your answer.

Industrial Production (a) Economists often use the expression expression "rate of growth" in relative rather than absolute terms. For example, let \(u=f(t)\) be the number of people in the labor force at time \(t\) in a given industry. (We treat this function as though it were differentiable even though it is an integer-valued step function.) Let \(v=g(t)\) be the average production per person in the labor force at time \(t .\) The total production is then \(y=u v\) . If the labor force is growing at the rate of 4\(\%\) per year year \((d u / d t=\) 0.04\(u\) ) and the production per worker is growing at the rate of 5\(\%\) per year \((d v / d t=0.05 v),\) find the rate of growth of the total production, y. (b) Suppose that the labor force in part (a) is decreasing at the rate of 2\(\%\) per year while the production per person is increasing at the rate of 3\(\%\) per year. Is the total production increasing, or is it decreasing, and at what rate?

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