Problem 6 - Applied Calculus

Question

Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ y=t^{2}(3 t+1)^{3} $$

Step-by-Step Solution

Answer

The derivative is \( (3t+1)^2(15t^2 + 2t) \).

Step 1: Recognize the Structure

The function given is a product of two functions: 1. \( u(t) = t^2 \)2. \( v(t) = (3t+1)^3 \)We will use the Product Rule to differentiate \( y = u(t) \, v(t) \). The Product Rule is \( (uv)' = u'v + uv' \).

Step 2: Differentiate \( u(t) = t^2 \)

Differentiate \( u(t) \) with respect to \( t \):\[ u'(t) = \frac{d}{dt}(t^2) = 2t \]

Step 3: Differentiate \( v(t) = (3t+1)^3 \)

Differentiate \( v(t) = (3t+1)^3 \) using the Chain Rule. Let \( g(t) = 3t+1 \) and \( v(t) = g(t)^3 \).Differentiating gives:\[ \frac{dv}{dt} = 3g(t)^2 \cdot g'(t) \]Since \( g(t) = 3t+1 \), we have \( g'(t) = 3 \).Thus, \[ v'(t) = 3(3t+1)^2 \cdot 3 = 9(3t+1)^2 \]

Step 4: Apply the Product Rule

Using the Product Rule, substitute the derivatives:\[ \frac{dy}{dt} = u'(t) v(t) + u(t) v'(t) \]Substituting \( u'(t) = 2t \), \( v(t) = (3t+1)^3 \), \( u(t) = t^2 \), and \( v'(t) = 9(3t+1)^2 \):\[ \frac{dy}{dt} = (2t)(3t+1)^3 + (t^2)(9)(3t+1)^2 \]

Step 5: Simplify the Expression

Simplify the expression:\[ 2t(3t+1)^3 + 9t^2(3t+1)^2 \]Factor out the common \( (3t+1)^2 \):\[ = (3t+1)^2 [2t(3t+1) + 9t^2] \]Expand and simplify:\[ = (3t+1)^2 [6t^2 + 2t + 9t^2] \]Combine like terms:\[ = (3t+1)^2 [15t^2 + 2t] \]So, \( \frac{dy}{dt} = (3t+1)^2(15t^2 + 2t) \).

Key Concepts

Product Rule

When dealing with the differentiation of a function that is a product of two separate functions, the product rule comes into play. It's a handy tool that helps you derive the derivative of the entire expression. Imagine you have a function, say, \( y = u(t) \, v(t) \), where both \( u(t) \) and \( v(t) \) are functions of \( t \). To differentiate this, we break it down using the product rule formula:

• \( (uv)' = u'v + uv' \)

This means you need to differentiate each function separately and then plug back into the formula. First, find the derivative of \( u(t) \) and multiply it by \( v(t) \), then find the derivative of \( v(t) \) and multiply by \( u(t) \), and finally, add both results together.
In our example problem, we have \( u(t) = t^2 \) and \( v(t) = (3t+1)^3 \). By applying the product rule, you'll notice the simplicity in handling even seemingly complex products of functions.

Chain Rule

The chain rule is crucial when you have a function nested within another function. Think of it like peeling an onion—each layer depends on the one above it.
If you encounter a composite function, such as \( v(t) = (3t+1)^3 \), you use the chain rule to differentiate it with respect to \( t \). Let's see how this works with the chain rule formula:

• \( \, d/dt \, [g(t)^n] = n \, g(t)^{n-1} \, g'(t) \)

Here, we consider \( g(t) = 3t + 1 \) as the inner function and \( g(t)^3 \) as the outer function.
By applying the chain rule, calculate the derivative of the outer function, multiply it by the derivative of the inner function, and you get the derivative of the composite function. It streamlines the process of breaking down more complex derivatives into manageable parts.

Differentiation

Differentiation is a fundamental principle in calculus, essentially helping us find the rate at which something is changing. When you differentiate a function, you are finding its derivative, which shows how the function value changes as its input changes.
Consider the function \( y = t^2 (3t+1)^3 \). Differentiating this function is not straightforward, since it's a combination of two functions, so you need tools like the product and chain rules. Each rule aids in differentiating specific parts of a complex function.
As shown in our example, we differentiate \( u(t) = t^2 \) simply by applying basic differentiation rules, resulting in \( u'(t) = 2t \). For the \( v(t) = (3t+1)^3 \) part, the chain rule steps in, as explained earlier. Both steps together demonstrate the utilization of differentiation in a practical setting.

Simplification

Once derivatives are calculated, simplification is often an essential final step. Simplification means making the expression more manageable or easier to work with. It's like tidying up a room; everything is in its place and easier to understand.
For the expression \( 2t(3t+1)^3 + 9t^2(3t+1)^2 \), a crucial part of simplification is factoring. By recognizing common factors, you reduce the expression to a simpler one that conveys the same information:

• Factor out \( (3t+1)^2 \) from each term: \( (3t + 1)^2 [2t(3t+1) + 9t^2] \)


• Then expand: \( (3t + 1)^2 [6t^2 + 2t + 9t^2] \)

After combining terms, you achieve the final, clean expression \( (3t+1)^2(15t^2 + 2t) \). This step highlights how simplification turns a raw mathematical expression into something more polished.