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Chapter 3: Problem 6

If \(x=y^{3}-y\) and \(d y / d t=5,\) then what is \(d x / d t\) when \(y=2 ?\)

Short Answer

Expert verified
When \( y=2 \), \( \frac{dx}{dt} = 55 \).

Step by step solution

01

Differentiate with respect to time

We start by differentiating both sides of the equation \( x = y^3 - y \) with respect to time \( t \). This gives us the expression for \( \frac{dx}{dt} \) in terms of \( y \), \( \frac{dy}{dt} \), and its derivatives. Using the chain rule, we get:\[ \frac{dx}{dt} = \frac{d}{dt}(y^3) - \frac{d}{dt}(y) \]\[ \frac{dx}{dt} = 3y^2 \frac{dy}{dt} - \frac{dy}{dt} \].
02

Substitute given values

We are given that \( \frac{dy}{dt} = 5 \) and we need to find \( \frac{dx}{dt} \) when \( y = 2 \). Substitute these values into the differentiated expression:\[ \frac{dx}{dt} = 3(2)^2 \times 5 - 5 \].
03

Simplify the expression

Simplify the expression obtained in the previous step:\[ \frac{dx}{dt} = 3 \times 4 \times 5 - 5 \]\[ \frac{dx}{dt} = 60 - 5 \]\[ \frac{dx}{dt} = 55 \].

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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 involves computing the derivative of a function. In simpler terms, a derivative measures how a function changes as its input changes. When differentiating a function, the rate of change is expressed with respect to a variable.
Here are some key points about differentiation:
  • A derivative provides the slope of the tangent line to the curve of a function at any given point.
  • Differentiation can be described as the process of finding the derivative, which indicates how the function's output value changes as the input value changes.
  • Notation wise, if we have a function \( f(x) \), then its derivative is often written as \( f'(x) \) or \( \frac{df}{dx} \).
  • In the given problem, when we differentiated the equation \( x = y^3 - y \) with respect to time, we were essentially calculating how \( x \) changes as both \( y \) and time \( t \) change.
Differentiation is a powerful tool because it allows us to understand dynamic systems and how they evolve over time or other parameters.
Chain Rule
The Chain Rule is a technique in calculus used to find the derivative of composite functions. This rule is essential when differentiating an expression where one variable depends on another, which in turn depends on a third variable.
Here's a simplified breakdown:
  • If we have an outside function and an inside function, the Chain Rule helps to differentiate them step by step.
  • Mathematically, if you have a function \( h(x) = f(g(x)) \), the derivative \( \frac{dh}{dx} \) is given by \( f'(g(x)) \times g'(x) \).
  • In our exercise, \( y \) is a function of \( t \) and we wanted to find how \( x = y^3 - y \) changes with respect to \( t \). Using the Chain Rule, we used \( y \) and \( \frac{dy}{dt} \) to find \( \frac{dx}{dt} \).
By using the Chain Rule, we effectively linked the rates of change of \( y \) and \( x \) with respect to time \( t \). This provided a methodical approach to solving problems with multiple interdependent variables.
Implicit Differentiation
Implicit Differentiation is a method used when the relationship between variables is expressed implicitly, not as a function explicitly solved for one variable. In these cases, it's not easy or possible to solve a variable directly in terms of others, and we resort to differentiating both sides of the equation as they are.
Here's the gist of implicit differentiation:
  • It allows us to find the derivative when functions are not explicitly separated.
  • You differentiate each term with respect to a common variable, often applying the Chain Rule as well.
  • When differentiating terms involving multiple variables, treat each variable as a function of the variable you are differentiating with respect to, using the process of implicit differentiation.
  • In the original exercise, we implicitly differentiated both sides of the equation \( x = y^3 - y \) with respect to \( t \), acknowledging that \( y \) is changing over time and due to its relationship with \( x \).
Implicit differentiation is particularly handy when dealing with equations where separating variables is complicated or unwieldy, streamlining the process of finding derivatives of complex relationships.

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Most popular questions from this chapter

Measuring acceleration of gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity g. The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in g. By keeping track of \(\Delta T,\) we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L .\) $$ \begin{array}{l}{\text { a. With } L \text { held constant and } g \text { as the independent variable, }} \\ {\text { calculate } d T \text { and use it to answer parts (b) and (c). }} \\ {\text { b. If } g \text { increases, will } T \text { increase or decrease? Will a pendulum }} \\ {\text { clock speed up or slow down? Explain. }}\end{array} $$ $$ \begin{array}{l}{\text { c. A clock with a } 100 \text { -cm pendulum is moved from a location }} \\ {\text { where } g=980 \mathrm{cm} / \mathrm{sec}^{2} \text { to a new location. This increases the }} \\ {\text { period by } d T=0.001 \mathrm{sec} . \text { Find } d g \text { and estimate the value of }} \\\ {g \text { at the new location. }}\end{array} $$

Unclogging arteries The formula \(V=k r^{4},\) discovered by the physiologist Jean Poiseuille \((1797-1869)\) , allows us to predict how much the radius of a partially clogged artery has to be expanded in order to restore normal blood flow. The formula says that the volume \(V\) of blood flowing through the artery in a unit of time at a fixed pressure is a constant \(k\) times the radius of the artery to the fourth power. How will a 10\(\%\) increase in \(r\) affect \(V\) ?

Motion in the plane The coordinates of a particle in the metric \(x y\)-plane are differentiable functions of time \(t\) with \(d x / d t=\) \(-1 \mathrm{m} / \mathrm{sec}\) and \(d y / d t=-5 \mathrm{m} / \mathrm{sec} .\) How fast is the particle's distance from the origin changing as it passes through the point \((5,12) ?\)

Highway patrol A highway patrol plane flies 3 mi above a level, straight road at a steady 120 \(\mathrm{mi} / \mathrm{h} .\) The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 \(\mathrm{mi}\) , the line-of-sight distance is decreasing at the rate of 160 \(\mathrm{mi} / \mathrm{h} .\) Find the car's speed along the highway.

Area Suppose that the radius \(r\) and area \(A=\pi r^{2}\) of a circle are differentiable functions of \(t\) . Write an equation that relates \(d A / d t\) to \(d r / d t .\)
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