Question

Using Related Rates In Exercises \(1-4,\) assume that \(x\) and \(y\) are both differentiable functions of \(t\) and find the required values of \(d y / d t\) and \(d x / d t .\) $$ \begin{array}{rlrl}{y=\sqrt{x}} & {\text { (a) } \frac{d y}{d t} \text { when } x=4} & {} & {\frac{d x}{d t}=3} \\ {} & {\text { (b) } \frac{d x}{d t} \text { when } x=25} & {} & {\frac{d y}{d t}=2}\end{array} $$

Short answer

The required values of \(dy/dt\) and \(dx/dt\) are 0.75 and 20 respectively.

Step-by-step solution

Differentiate the given function

The given function is \(y = \sqrt{x}\). This can also be written as \(y = x^{1/2}\). So, when we differentiate it with respect to \(t\), we will use the chain rule. Differentiating \(y\) gives \(dy/dt = 1/2 * x^{-1/2} * dx/dt = dx/dt / (2\sqrt{x})\).

Find dy/dt when x=4 and dx/dt=3

To find \(dy/dt\) when \(x=4\) and \(dx/dt=3\), substitute \(x=4\) and \(dx/dt=3\) into the derivative equation. So, \(dy/dt = 3 / (2 * \sqrt{4}) = 3/4 = 0.75\).

Find dx/dt when x=25 and dy/dt=2

We rearrange the equation for \(dx/dt\). So, \(dx/dt = 2y * dy/dt\). To find \(dx/dt\) when \(x=25\) and \(dy/dt=2\), substitute \(x=25\) and \(dy/dt=2\) into the equation, remembering that \(y = \sqrt{x}\) implies \(y = 5\) when \(x = 25\). So, \(dx/dt = 2*5*2 = 20\).

Key concepts

Chain Rule

The chain rule is a fundamental tool in calculus used when taking the derivative of a composite function. Essentially, it helps us understand how changes in one quantity affect another interconnected quantity. For instance, if you have a function nested within another, such as when you differentiate expressions like \(y = \sqrt{x} = x^{1/2}\), the chain rule allows you to compute the rate of change with respect to time, or any other variable like \(t\).
To apply the chain rule to \(y = x^{1/2}\), differentiate the outer function \(f(u) = u^{1/2}\) with respect to \(u\) and then multiply by the derivative of the inner function \(u = x\) with respect to \(x\). This gives us: \[\frac{dy}{dt} = \frac{1}{2} x^{-1/2} \cdot \frac{dx}{dt}, \]

• The derivative \(\frac{1}{2} x^{-1/2}\) corresponds to the derivative of \(x^{1/2}\).
• The \(\frac{dx}{dt}\) represents the rate of change of \(x\) over time.

The chain rule is vital in scenarios where variables are interdependent, allowing us to express complicated scenarios in simpler terms.

Differentiable Functions

A differentiable function is a function that has a derivative at each point in its domain. In simpler words, if a function can be represented by a smooth line without any breaks, sharp corners, or cusps, it is differentiable.
In our exercise, both \(x\) and \(y\) are stated to be differentiable functions of \(t\), which implies:

• Both quantities change smoothly over time.
• We can compute their rates of change using derivatives.

The derivative, which represents the rate of change or slope of the function, is the cornerstone of understanding how quantitative relationships evolve. It is essential for related rates problems, as these typically involve finding multiple interrelated derivatives with respect to a third variable, in this case, \(t\).
When a function is differentiable, like \(y=\sqrt{x}\) in our problem, we can use various calculus tools, such as the chain rule, to explore its properties and gather useful insights about how the function behaves over time.

Derivatives

Derivatives provide us with a way to understand the rate at which one quantity changes with respect to another. In the realm of calculus, the process of differentiation gives us the derivative of a function, which captures this rate of change in mathematical terms.
In our specific problem, the task is to find the derivatives \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), where these derivatives show:

• \(\frac{dy}{dt}\): How \(y\) changes as time \(t\) progresses.
• \(\frac{dx}{dt}\): How \(x\) changes over the same time interval.

The derivative of \(y = \sqrt{x}\) is particularly straightforward when utilizing the chain rule. Differentiating gives \(\frac{dy}{dx} = \frac{1}{2} x^{-1/2}\), and further, through the chain rule, the time derivative becomes \(\frac{dy}{dt} = \frac{1}{2} x^{-1/2} \cdot \frac{dx}{dt}\).
Every instance where you compute a derivative opens a portal into understanding dynamics thoroughly—be it the speed of a car or the growth rate of an investment. This powerful tool is invaluable across various fields of science and engineering, providing clarity in measuring and predicting changes.

Implicit Differentiation

Implicit differentiation is a useful technique for finding derivatives of equations that define relationships between variables in an implicit manner, rather than as explicit functions. In such scenarios, one variable is not isolated on one side of the equation.
For the equation \(y = \sqrt{x}\), which can be written as \(y^2 = x\), both variables \(x\) and \(y\) can be differentiated with respect to \(t\) by using implicit differentiation. In contrast to explicit differentiation, we treat both \(x\) and \(y\) as functions of \(t\) and differentiate both sides of the equation separately with respect to \(t\):

• If \(y = x^{1/2}\), we directly apply the chain rule to find \(\frac{dy}{dt}\) in terms of \(\frac{dx}{dt}\).
• If you rearrange to \(x = y^2\), this relation leads to \(\frac{dx}{dt} = 2y \cdot \frac{dy}{dt}\).

Implicit differentiation shines where it might be complex or impossible to isolate a given variable. By directly differentiating each variable with respect to a common parameter, it simplifies the computation of derivatives in an otherwise complicated setup of equations like in related rates scenarios.