Question

Assume that \(x\) and \(y\) are both differentiable functions of \(t\) and find the required values of \(d y / d t\) and \(d x / d t\) $$ \begin{array}{llll} \underline{\text { Equation }} & \text { Find } & \text { Given } \\ \hline y=\sqrt{x} & \text { (a) } \frac{d y}{d t} \text { when } x=4 & \frac{d x}{d t}=3 \\ & \text { (b) } \frac{d x}{d t} \text { when } x=25 & \frac{d y}{d t}=2 \end{array} $$

Short answer

The value of \(dy/dt\) when \(x = 4\) and \(dx/dt = 3\) is \(\frac{3}{4}\). The value of \(dx/dt\) when \(x = 25\) and \(dy/dt = 2\) is 20.

Step-by-step solution

Differentiation

Differentiating \(y = \sqrt{x}\) with respect to \(t\) using chain rule gives us \(\frac{dy}{dt} = \frac{1}{2 \sqrt{x}}\cdot\frac{dx}{dt}\).

Finding \(dy/dt\) when \(x=4\) given \(dx/dt=3\)

Substitute \(x=4\) and \(dx/dt=3\) into the equation: \(\frac{dy}{dt} = \frac{1}{2 \sqrt{4}}\cdot3 = \frac{3}{4}\).

Finding \(dx/dt\) when \(x = 25\) given \(dy/dt=2\)

Rearranging our equation to find \(dx/dt\) gives us \(\frac{dx}{dt} = 2\sqrt{x} \cdot \frac{dy}{dt}\). Substituting \(x = 25\) and \(dy/dt = 2\) into the equation gives us: \(\frac{dx}{dt} = 2\sqrt{25} \cdot 2 = 20\).

Key concepts

Chain Rule Differentiation

Understanding the chain rule is crucial in calculus, especially when dealing with composite functions where one variable is a function of another variable, which in turn, is a function of a third variable. Essentially, it's about how to differentiate a function of a function. If we consider an equation like y = \(\sqrt{x}\), where both x and y are functions of t, we need to differentiate y with respect to t through x. This is where the chain rule comes in to accommodate the indirect relationship.

The chain rule states that the derivative of y with respect to t is the derivative of y with respect to x multiplied by the derivative of x with respect to t, or \(\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}\). In our example, after applying the chain rule, we find that \(\frac{dy}{dt} = \frac{1}{2\sqrt{x}}\cdot\frac{dx}{dt}\).

When applying the chain rule, it's essential to differentiate each part accurately and not to confuse the order of multiplication. As seen in the exercise, even though the square root function might initially seem intimidating, it's quite straightforward with the chain rule.

Implicit Differentiation

Implicit differentiation comes into play when it's either impossible or impractical to solve for one variable in terms of another explicitly. Most of the functions we encounter are explicitly defined, for example, \(y = x^2 + 3\). However, not all relationships between x and y are this straightforward. Implicit differentiation allows us to differentiate equations where y is not isolated on one side of the equation.

For implicit differentiation, we treat y as an implicit function of x and differentiate both sides of the equation with respect to x, keeping in mind the chain rule when differentiating terms that include y. This method can yield expressions for \(\frac{dy}{dx}\) that can then be evaluated given certain conditions or additional information, such as the value of \(\frac{dx}{dt}\), as in the given exercise.

Related Rates Calculus

The concept of related rates allows us to study the relationship between rates at which different quantities change. In these problems, we are typically given the rate of change of one quantity and asked to find the rate of change of another quantity to which it is related, usually through an equation or a geometric shape.

In the exercise provided, we see this concept in action. We're given the rate of change of x, that is \(\frac{dx}{dt}\), and we are required to find the rate of change of y. Here's how we approach it: Given our equation y = \sqrt{x}, we take the derivative with respect to t applying chain rule, and then with the known rate of the change of x, we can find the rate of change of y. This technique is powerful in many real-life applications ranging from physics to economics, where multiple variables are interdependent, and understanding the rate at which one changes is pivotal for deducing information about another.