Problem 1 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(g(t)=3^{5 t} .\) Show that $$ \frac{g(t+h)-g(t)}{h}=g(t) \cdot \frac{g(h)-g(0)}{h} . $$

Step-by-Step Solution

Answer

Yes, the given equation is indeed true. By leveraging the properties of exponents and making appropriate substitutions, we found that both sides of the equation are identical.

Step 1: Analyze the given function

The given function is \(g(t) = 3^{5t}\). It is an exponential function. As for any exponential function, the base raised to the exponent of the sum of values is equal to the product of the bases raised to these values. This property will be very helpful in the next steps.

Step 2: Express \(g(t+h)\) using properties of exponentials

Expand \(g(t+h) = 3^{5t+5h} = 3^{5t} \cdot 3^{5h} = g(t) \cdot g(h)\). This shows that separating the terms in the exponent allows us to express \(g(t+h)\) in terms of \(g(t)\) and \(g(h)\), using the multiplication property of exponents.

Step 3: Likewise express \(g(h)-g(0)\)

Aggregate \(g(h) - g(0)\) similarly resulting \(g(h) - g(0) = 3^{5h} - 3^{5 \cdot 0} = g(h) - 1\). At \(t = 0\), the function \(g(t)\) has a value of 1, as any number (except 0) raised to zero equals 1. Therefore, \(g(0) = 1\).

Step 4: Substitute expressions from Steps 2 and 3 into the equations

Now, substitute \(g(t+h)\) and \(g(h)-g(0)\) we just found into the equation. This gives us: \[\frac{g(t) \cdot g(h) - g(t)}{h} = g(t) \cdot \frac{g(h) - 1}{h}\] Now distribute \(g(t)\) through the right side of equation: \[= g(t) \cdot \frac{g(h)}{h} - g(t)\]. Notice the left side and right side are the same, which verifies the equation.

Key Concepts

Properties of Exponents

When dealing with exponential functions, understanding the properties of exponents is crucial. An exponential function is generally of the form \(a^x\) where \(a\) is a constant and \(x\) is the exponent. Here are some key properties that apply:

• The Product of Powers: When multiplying like bases, add the exponents: \(a^m \cdot a^n = a^{m+n}\).
• The Power of a Power: When raising a power to another power, multiply the exponents: \((a^m)^n = a^{mn}\).
• The Power of a Product: Distribute the exponent to each factor: \((ab)^m = a^m b^m\).
• The Zero Exponent: Any base except zero raised to the power of zero is 1: \(a^0 = 1\).

These properties are especially useful in simplifying the expressions in exponential form and are applied heavily in calculus, including in our exercise example. For instance, understanding how \(g(t+h) = 3^{5t} \cdot 3^{5h}\) simplified to \(g(t) \cdot g(h)\) in our problem relies directly on the product of powers property.

Limit of a Function

The concept of a limit is foundational in calculus and relates to how a function behaves as its input approaches some value. To understand the limit of a function, envision it as determining the value that the function approaches as the input gets arbitrarily close to a specific point. Limits help define critical calculus concepts, such as:

• Continuity: A function is continuous at a point if the limit exists at that point and equals the function's value.
• Derivatives: The basis for understanding rates of change. The derivative at a point is the limit of the difference quotient as the increment approaches zero.

To use limits effectively, especially in problems like ours, it’s often necessary to simplify expressions using algebra, including properties of exponents, which can be crucial in finding the limit accurately. For example, in our given exercise, the steps of simplifying expressions with exponential functions are pivotal in determining their limits correctly.

Difference Quotient

The difference quotient of a function is a fundamental tool in calculus that allows us to approximate the derivative or the slope of the tangent at a point. It is formed by the expression:\[ \frac{f(x+h) - f(x)}{h}. \]This formula finds the average rate of change of the function over the interval \(h\). As \(h\) approaches zero, it gets us closer to finding the instantaneous rate of change, or the derivative. Here's how it works in the context of our problem:- First, substitute the expressions for \(g(t+h)\) and \(g(t)\).- By simplifying the expression using properties of exponents, observe that our transformation holds true.In our scenario with \(g(t) = 3^{5t}\), expressing it in terms of \(g(t)\) and \(g(h)\) like \(g(t+h) = g(t) \cdot g(h)\) and simplifying \(g(h) - g(0)\) to \(g(h) - 1\) makes it evident how the difference quotient emerges naturally.Understanding and manipulating the difference quotient helps in grasping fundamental calculus concepts like derivatives and in mastering how to handle various types of functions efficiently.