Question

Moving Point In Exercises \(5-8,\) a point is moving along the graph of the given function at the rate \(d x / d t .\) Find \(d y / d t\) for the given values of \(x .\) $$ \begin{array}{l}{y=\frac{1}{1+x^{2}} ; \frac{d x}{d t}=6 \text { inches per second }} \\ {\begin{array}{ll}{\text { (a) } x=-2} & {\text { (b) } x=0} & {\text { (c) } x=2}\end{array}}\end{array} $$

Short answer

(a) For \(x=-2\), the rate of change of \(y\) is \(-3/5\) inches/second. (b) For \(x=0\), the rate of change of \(y\) is \(0\) inches/second. (c) For \(x=2\), the rate of change of \(y\) is \(-3/5\) inches/second.

Step-by-step solution

Compute dy/dx

We begin by computing the derivative of \(y\) with respect to \(x\). We use the quotient rule of differentiation to find that \(dy/dx = -2x/(1+x^2)^2\). In our case, the function is \(y = 1/(1 + x^{2})\). We can represent this as a fraction, so the numerator (\(u\)) is 1 and the denominator (\(v\)) is \(1+x^{2}\). Differentiating them separately gives us \(u'=0\) and \(v'= 2x\), therefore, differentiating \(y\) by the quotient rule gives \(dy/dx = -(v.u' - u.v') / v^{2} = -2x / (1+x^{2})^{2}\).

Compute dy/dt for dx/dt = 6

Using the chain rule of differentiation, we know that \(dy/dt = dy/dx * dx/dt\). Here, dx/dt is given as 6, so substituting this value along with dy/dx, which we found in step 1, in the formula gives us \(dy/dt = 6*(-2x/(1+x^2)^2)\) , a formula to compute \(dy/dt\) at any value of \(x\).

Compute dy/dt for specific x values

Now, let's substitute our specific \(x\) values: -2, 0, and 2 to get the corresponding dy/dt values. (a) For \(x=-2\), \(dy/dt = 6*2*(-2)/(1 + (-2)^{2})^{2} = -3/5 inches/second. (b) For \(x=0\), there is no dependence on \(x\) in the function so dy/dt will be zero. (c) For \(x=2\), \(dy/dt = 6*2*2 /(1+ 2^{2})^{2}= -3/5 inches/second. Please note that the unit for dy/dt will be inches/second because dx/dt was given in inches/second.

Key concepts

Chain Rule of Differentiation

The chain rule is a fundamental tool in calculus, particularly useful in related rates problems. It allows us to determine the rate at which one variable changes in relation to another. Imagine you have a chain of functions, one inside the other, like a Russian nesting doll. The chain rule helps you find the derivative of this complex function by multiplying the derivatives of each nested function, link by link.

When applied to related rates, the chain rule tells us that to find the rate of change of one quantity with respect to time, we can multiply the rate of change of that quantity with respect to another variable by the rate of change of that second variable with respect to time. Mathematically, if you have a dependent variable y, which relies on an independent variable x, which itself changes over time, the chain rule is expressed as: \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \).

For example, if we know the speed at which x changes over time (\(\frac{dx}{dt}\)) and the relationship between y and x (\(\frac{dy}{dx}\)), the chain rule allows us to calculate the rate at which y changes over time (\(\frac{dy}{dt}\)) with ease. This concept was crucial to determine \(\frac{dy}{dt}\) for the provided exercise, where the movement of a point along a function's graph was studied.

Quotient Rule of Differentiation

The quotient rule is another essential differentiation technique, especially when dealing with functions that are fractions of one another. It's the strategy to use whenever you're differentiating a function that is the quotient of two other functions. Its formula is intuitive if you think of it in terms of the high school mnemonic: 'low d-high minus high d-low, square the bottom, and away we go'.

In more formal terms, for functions u(x) and v(x), where y = u/v, the quotient rule states that \( \frac{dy}{dx} = \frac{v \cdot u' - u \cdot v'}{v^2} \). Here, \(u'\) and \(v'\) are the derivatives of u and v with respect to x, respectively.

In the context of our exercise, where \(y = \frac{1}{1+x^{2}}\), we defined u as 1 and v as \((1+x^{2})\). The quotient rule helped us find \(\frac{dy}{dx}\), which is essential in further steps to determine \(\frac{dy}{dt}\) using the chain rule.

Rate of Change

The rate of change is a measure of how a quantity y grows or shrinks in relation to another quantity x. When x represents time, it's the speed at which y is changing as time advances. It's what most people commonly think of as the 'speed' of an object or the 'growth rate' of a plant.

In calculus, the rate of change is expressed as a derivative, indicating it is about the instantaneous change at a certain point, rather than an average over an interval. If you're calculating how fast a car is moving at a specific instant, you're thinking about its rate of change of position with respect to time.

From our exercise, we are concerned with two rates: the rate at which x changes with time (\(\frac{dx}{dt}\)), and the rate at which y changes with time (\(\frac{dy}{dt}\)). By understanding both these rates and applying the chain and quotient rules, we can effectively tackle related rates problems in calculus.