Problem 65 - Calculus and its applications

Question

Differentiate each function. \(y(t)=5 t(t-1)(2 t+3)\)

Step-by-Step Solution

Answer

The derivative of the function is \( y'(t) = 30t^2 + 10t - 15 \).

Step 1: Apply the Product Rule

The function is a product of three functions: \( u(t) = 5t \), \( v(t) = (t-1) \), and \( w(t) = (2t+3) \). We will use the product rule for three functions: \( y'(t) = u'(t)v(t)w(t) + u(t)v'(t)w(t) + u(t)v(t)w'(t) \).

Step 2: Differentiate First Function

Differentiate \( u(t) = 5t \) to get \( u'(t) = 5 \).

Step 3: Differentiate Second Function

Differentiate \( v(t) = (t-1) \) to get \( v'(t) = 1 \).

Step 4: Differentiate Third Function

Differentiate \( w(t) = (2t+3) \) to get \( w'(t) = 2 \).

Step 5: Apply Derivatives to Product Rule

Now substitute the derivatives into the product rule formula: \[ y'(t) = 5(t-1)(2t+3) + 5t(1)(2t+3) + 5t(t-1)2 \].

Step 6: Simplify Each Term

Simplify each term: 1. \( 5(t-1)(2t+3) = 5(2t^2 + 3t - 2t - 3) = 10t^2 + 5t - 15 \)2. \( 5t(2t+3) = 10t^2 + 15t \)3. \( 5t(t-1)2 = 10t^2 - 10t \)

Step 7: Combine All Terms

Add all the simplified terms together: \( y'(t) = (10t^2 + 5t - 15) + (10t^2 + 15t) + (10t^2 - 10t) \).

Step 8: Further Simplification

Combine like terms to yield the final answer: \( y'(t) = 30t^2 + 10t - 15 \).

Key Concepts

Product Rule

When dealing with the differentiation of a product of functions, the product rule is essential. The product rule gives us a method to differentiate the product of two or more functions. In the context of this exercise, we have the function \( y(t) = 5t(t-1)(2t+3) \), which is a product of three separate functions: \( u(t) = 5t \), \( v(t) = (t-1) \), and \( w(t) = (2t+3) \).
You might wonder how to handle a product of three functions since we usually see the product rule applied to only two. The general formula adapts easily by differentiating one function at a time while keeping the rest unchanged in their turn. This results in the product rule for three functions:

• \( y'(t) = u'(t)v(t)w(t) + u(t)v'(t)w(t) + u(t)v(t)w'(t) \)

This approach helps in systematically breaking down the problem and makes it easier to manage the derivatives of each separate portion of the function.

Derivative

Differentiation is a core concept of calculus that allows us to find the rate at which a function is changing at any point. This rate of change is what's known as the derivative of the function. In the original exercise, we were required to find \( y'(t) \), which represents the derivative of the function \( y(t) = 5t(t-1)(2t+3) \).
The steps begin by differentiating each part of the product individually:

• The derivative of \( u(t) = 5t \) is \( u'(t) = 5 \).
• The derivative of \( v(t) = (t-1) \) is \( v'(t) = 1 \).
• The derivative of \( w(t) = (2t+3) \) is \( w'(t) = 2 \).

Each of these partial derivatives is then substituted back into the product rule formula, allowing us to systematically construct the overall derivative, \( y'(t) \), of the original function. Differentiation lays the foundation for understanding and solving more complex problems in calculus.

Polynomial Function

A polynomial function is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. The function in the exercise \( y(t) = 5t(t-1)(2t+3) \) can be expanded and simplified to give a polynomial expression after using the product rule and simplifying the derivatives.
Polynomial functions have coefficients, exponents, and they can be a useful model for various real-world situations due to their general form and flexibility. The result of this exercise, after all terms have been combined and simplified, is the polynomial function \( y'(t) = 30t^2 + 10t - 15 \).
This expression is a second-degree polynomial, showing highest exponent of \( t \) as 2. Navigating through polynomial manipulation requires strong algebraic skills, and understanding polynomial functions is essential for analysing the behavior of graphs and solving equations.