Problem 63 - Calculus for Biology and Medicine

Question

Assume that \(x\) and \(y\) are differentiable functions of \(t\). Find \(\frac{d y}{d t}\) when \(x^{2} y=1\) and \(\frac{d x}{d t}=3\) for \(x=2\).

Step-by-Step Solution

Answer

\(\frac{dy}{dt} = -\frac{3}{4}\) when \(x = 2\).

Step 1: Differentiate the Given Equation

Start by differentiating the equation \(x^2y = 1\) with respect to \(t\). Use the product rule for differentiation since both \(x\) and \(y\) are functions of \(t\). The product rule states that for two functions \(u\) and \(v\), \(\frac{d}{dt}(uv) = u'v + uv'\). Here, set \(u = x^2\) and \(v = y\).

Step 2: Apply the Product Rule

Using the product rule and chain rule, differentiate: \[ \frac{d}{dt}(x^2y) = \frac{d}{dt}(x^2) \cdot y + x^2 \cdot \frac{dy}{dt}\]. Now differentiate \(x^2\) which gives \(2x \cdot \frac{dx}{dt}\). So, substitute to get: \[ 2x \frac{dx}{dt} \cdot y + x^2 \cdot \frac{dy}{dt} = 0\].

Step 3: Substitute Given Values

Substitute the given values into the differentiated equation: \(x = 2\) and \(\frac{dx}{dt} = 3\). The equation becomes: \[ 2(2)(3) \cdot y + (2)^2 \cdot \frac{dy}{dt} = 0 \]. Simplify this to: \[ 12y + 4 \frac{dy}{dt} = 0 \].

Step 4: Solve for \(\frac{dy}{dt}\)

Re-arrange the equation to solve for \(\frac{dy}{dt}\): \[ 4 \frac{dy}{dt} = -12y \]. Simplify to: \[ \frac{dy}{dt} = -3y \].

Step 5: Find the Value of \(y\) at \(x = 2\)

Given \(x^2y = 1\), substitute \(x = 2\) to find \(y\): \[ (2)^2y = 1 \Rightarrow 4y = 1 \Rightarrow y = \frac{1}{4} \].

Step 6: Substitute \(y\) Back to Determine \(\frac{dy}{dt}\)

Substitute \(y = \frac{1}{4}\) back into the equation \(\frac{dy}{dt} = -3y\) to find \(\frac{dy}{dt}\): \[ \frac{dy}{dt} = -3 \times \frac{1}{4} = -\frac{3}{4} \].

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus for finding the derivative of a product of two functions. It allows us to compute the derivative of two differentiable functions multiplied together. Here's how it works:

• If you have two functions, say \( u \) and \( v \), then the derivative of their product \( uv \) with respect to a variable, say \( t \), is given by the formula: \[ \frac{d}{dt}(uv) = u'v + uv' \]
• In simple terms, you differentiate one function while keeping the other function the same, and then switch – differentiate the second function while keeping the first function the same.
• This rule is particularly useful when both functions depend on a common variable, as with compound functions like \( x^2 y \) in the given exercise.

In our exercise, we used the product rule to differentiate \( x^2 y \) because both \( x \) and \( y \) change with respect to \( t \). By setting \( u = x^2 \) and \( v = y \), we applied the product rule to obtain: \[ \frac{d}{dt}(x^2) \cdot y + x^2 \cdot \frac{dy}{dt} \] This step is crucial for solving the problem and shows the power of the product rule in handling multiple functions at once.

Chain Rule

The chain rule is another critical concept in calculus used for finding the derivative of a composite function. In simple terms, it helps us differentiate when a function is nested within another function. Here's what you need to know:

• For a composite function like \( f(g(t)) \), the chain rule states that the derivative is: \[ \frac{d}{dt}f(g(t)) = f'(g(t)) \, g'(t) \]
• It involves taking the derivative of the outer function, \( f \), at the inner function \( g(t) \), and then multiplying it by the derivative of the inner function \( g(t) \) with respect to \( t \).
• This method is handy when you're working with functions that depend on another function, as seen in our differentiation of \( x^2 \), where the differentiation involves a further function of \( t \).

In the context of our problem, the chain rule assists when differentiating \( x^2 \) with respect to \( t \). Specifically, you recognize that any differentiation of \( x \) concerning \( t \) requires applying the chain rule to factor in how \( x \) changes with \( t \). Thus, when we find \( \frac{d}{dt}(x^2) \), we use:\[ 2x \cdot \frac{dx}{dt} \]This application is crucial when variables themselves are functions rather than constants.

Differentiable Functions

Differentiable functions are essential in calculus, particularly for applying rules like the product rule and chain rule. Here’s a closer look:

• A function is said to be differentiable if it has a derivative at every point in its domain. This means the function's graph has no breaks, sharp corners, or vertical tangents.
• Differentiability is essential because it allows the use of calculus tools, like the product and chain rules, to find rates of change and solve more complex problems.
• In practical terms, for two functions \( x(t) \) and \( y(t) \) to be differentiable, as stated in the exercise, they must be smooth and continuous.

In our problem, assuming \( x \) and \( y \) are differentiable functions of \( t \) means that they smoothly change over time. This is a prerequisite for applying the differentiation rules effectively. Without differentiability, we couldn't confidently use either the product rule or chain rule, as these require the functions involved to be differentiable across their domains. This smooth behavior ensures precise calculations and reliable solutions to dynamically changing conditions.