Problem 81 - Calculus for Biology and Medicine

Question

Differentiate $$ h(t)=\sqrt{a t}(t-a)+a t $$ with respect to \(t\). Assume that \(a\) is a positive constant.

Step-by-Step Solution

Answer

The derivative is \( (at)^{1/2} + \left(\frac{a}{2}t - \frac{a^2}{2}\right)(at)^{-1/2} + a. \)

Step 1: Expand the Expression

First, expand the term \( \sqrt{a t}(t-a) \). This gives us: \[ \sqrt{a t} \times t - \sqrt{a t} \times a = t \sqrt{a t} - a \sqrt{a t}. \] Thus, the entire function \( h(t) \) can be rewritten as: \[ h(t) = t \sqrt{a t} - a \sqrt{a t} + a t. \]

Step 2: Differentiate Each Term Separately

Next, differentiate each term of \( h(t) = t \sqrt{a t} - a \sqrt{a t} + a t \) with respect to \( t \). 1. Differentiate \( t \sqrt{a t} \): - Rewrite \( \sqrt{a t} \) as \( (at)^{1/2} \). Use the product rule: \[ \frac{d}{dt}(t(at)^{1/2}) = \frac{d}{dt}(t) \times (at)^{1/2} + t \times \frac{d}{dt}((at)^{1/2}). \] - The derivative of \(t\) is 1, so the first term becomes \((at)^{1/2}\). - For the second term, use the chain rule on \((at)^{1/2}\): \[ \frac{1}{2} (at)^{-1/2} \times a = \frac{a}{2} (at)^{-1/2}. \] - Thus, the second term becomes \( \frac{a t}{2} (at)^{-1/2}\). - Overall derivative: \[ (at)^{1/2} + \frac{a t}{2} (at)^{-1/2}. \] 2. Differentiate \(-a \sqrt{a t}\): - This is a constant multiplied by \( (at)^{1/2} \), apply the chain rule: \[ -a \times \frac{1}{2} (at)^{-1/2} \times a = -\frac{a^2}{2} (at)^{-1/2}. \] 3. Differentiate \( a t \): - The derivative is simply \( a \), since \(a\) is a constant.

Step 3: Combine Differentiated Terms

Now, sum up all the differentiated parts for the function: \[ \frac{d}{dt} h(t) = (at)^{1/2} + \frac{a}{2} t (at)^{-1/2} - \frac{a^2}{2} (at)^{-1/2} + a. \] Simplify by combining terms involving \((at)^{-1/2}\): \[(at)^{1/2} + \left(\frac{a}{2}t - \frac{a^2}{2}\right)(at)^{-1/2} + a.\]

Key Concepts

Understanding the Product Rule

The product rule is a fundamental principle in calculus used to differentiate products of two functions. When differentiating the product of two functions, say \( u(t) \) and \( v(t) \), the product rule states that the derivative is:

• \( u'(t) \times v(t) + u(t) \times v'(t) \).

In the original exercise, we needed to differentiate the term \( t \sqrt{at} \). Here, \( t \) and \( (at)^{1/2} \) are the functions to apply the product rule to:

• Differentiate \( t \) to get 1.
• Leave \( (at)^{1/2} \) as it is for the first part.
• Then, differentiate \( (at)^{1/2} \) using the chain rule for the second part.

This method allows us to handle more complex products of functions efficiently. It emphasizes breaking down each component step by step to ensure clarity and accuracy during differentiation. Keep this rule in mind as it's essential for tackling problems involving the multiplication of functions.

Mastering the Chain Rule

The chain rule is another crucial tool in differentiation that helps us find the derivative of composite functions. Suppose we have a composite function \( f(g(x)) \), where \( f \) is a function of \( g \), which in turn is a function of \( x \). The chain rule states:

• The derivative of \( f(g(x)) \) is \( f'(g(x)) \times g'(x) \).

In the given exercise, we used the chain rule to differentiate \( (at)^{1/2} \). This term is a composite because it involves the function \( t \) inside the root:

• Differentiating the outer power \( (at)^{1/2} \) gives us \( \frac{1}{2} (at)^{-1/2} \).
• Then, we multiply by the derivative of the inner function \( at \), which is \( a \).

Using the chain rule effectively requires identifying the outer and inner functions, then carefully applying the rule while keeping track of each step. It simplifies the process of differentiation for nested or layered functions.

Applying Algebraic Manipulation

Algebraic manipulation is a technique used to simplify expressions, making them easier to work with or differentiate. In the original exercise, we started by expanding \( \sqrt{at}(t-a) \) into its individual components, resulting in:

• \( t \sqrt{at} - a \sqrt{at} \).

This expansion allowed us to clearly separate the terms for individual differentiation. Furthermore, combining like terms was necessary after differentiating:

• This involved summing the derivatives of each part, especially terms involving \( (at)^{-1/2} \).
• The goal is to rearrange and combine them to achieve the simplest form possible.

By employing algebraic manipulation, you'll find it simpler to tackle intricate expressions and differentiate them systematically. It goes hand-in-hand with other differentiation techniques to streamline these processes, providing a solid foundation for more complex calculations you might encounter.