Problem 81 - Calculus for Biology and Medicine

Question

Differentiate $$ h(t)=\sqrt{a t}(t-a)+a t $$

Step-by-Step Solution

Answer

The derivative of the function is \( h'(t) = \frac{a(t-a)}{2\sqrt{a t}} + \sqrt{a t} + a \).

Step 1: Identify the Function to Differentiate

The given function is \( h(t) = \sqrt{a t}(t-a) + a t \). This is comprised of two terms: \( \sqrt{a t}(t-a) \) and \( a t \). We need to apply differentiation rules to each term.

Step 2: Differentiate the First Term

The first term is \( \sqrt{a t}(t-a) \). Reexpress it as \((a t)^{1/2} (t-a)\) and apply the product rule: \( \frac{d}{dt}[uv] = u'v + uv' \), where \( u = (a t)^{1/2} \) and \( v = (t-a) \).

Step 3: Differentiate Each Part of the First Term

Find \( u' \): \( u = (a t)^{1/2} = (a t)^{0.5} \), so \( u' = 0.5 (a t)^{-0.5} a = \frac{a}{2\sqrt{a t}} \). Find \( v' \): \( v = (t-a) \), so \( v' = 1 \).

Step 4: Apply Product Rule to the First Term

Using the product rule: \( \frac{d}{dt}[(a t)^{1/2}(t-a)] = \frac{a}{2\sqrt{a t}}(t-a) + (a t)^{1/2}(1) \). Simplify this expression.

Step 5: Differentiate the Second Term

The second term is \( a t \). The derivative of \( a t \) with respect to \( t \) is \( a \), since \( a \) is a constant.

Step 6: Combine the Results

Add the derivatives of both terms: \( \frac{a}{2\sqrt{a t}}(t-a) + (a t)^{1/2} + a \). This is the derivative \( h'(t) \).

Step 7: Final Simplification

Express \( h'(t) = \frac{a(t-a)}{2\sqrt{a t}} + \sqrt{a t} + a \) in the simplest form possible. The expression is already in a simplified form, representing \( h'(t) \).

Key Concepts

Understanding the Product Rule

When differentiating a function that is the product of two or more functions, the product rule is indispensable. The product rule is a formula used to find the derivative of a product of two functions. Let's say we have two functions, \( u(t) \) and \( v(t) \). Their product's derivative can be written as:

• \( \frac{d}{dt}[uv] = u'v + uv' \)

To apply it effectively, break down the original functions and assign them to \( u \) and \( v \). Then, find each derivative, \( u' \) and \( v' \), separately. The sum of \( u'v \) and \( uv' \) will give you the derivative of the product.
In the context of this exercise, the first term is \( \sqrt{a t}(t-a) \). Here, \( u = (a t)^{1/2} \) and \( v = (t-a) \). Using the product rule, we're able to determine how these variables interact when differentiated, which simplifies the complex product into manageable steps. This method is a fundamental tool for handling products in calculus.

Grasping Derivatives

Derivatives measure how a function changes as its input changes. In simple terms, they describe the function's rate of change. When calculating a derivative, we look for the slope of the tangent line to the curve of a function at a given point. This concept is the backbone of differential calculus.
For functions involving constants and variables, such as \( a t \) in this exercise, the derivative involving a constant is straightforward. The derivative of \( a t \) with respect to \( t \) is simply \( a \) since \( t \) is differentiated as 1 and \( a \) is a constant multiplier.

• When handling more complex terms, like \( \sqrt{a t} \), revisiting rules for exponentials can be beneficial. Recall that \( \sqrt{a t} = (a t)^{1/2} \) which simplifies finding the derivative by applying power and chain rules effectively.

Ultimately, understanding derivatives is about grasping how each part of a function responds to changes, streamlining problems even when they initially seem daunting.

Mastering Simplification

After differentiating, the challenge often lies in simplifying the expression to its cleanest form. Without simplification, the derivative can remain complex and hard to interpret.
Simplification typically involves combining like terms, canceling common factors, and reducing fractions. In this exercise, after applying the product rule and differentiating each term, ensure each part of the derivative connects correctly.

• The expression \( \frac{a(t-a)}{2\sqrt{a t}} + \sqrt{a t} + a \) follows derivative rules but can be evaluated further for simplicity and conciseness.

Through these steps, expressions become clearer and show critical points of the original function more directly. Practicing simplification creates cleaner solutions and enhances understanding of a function's behavior at different points.