Problem 63 - Calculus - AP Edition

Question

A stuntman estimates the time \(T\) in seconds for him to fall \(x\) meters by \(T=0.453 \sqrt{x}\). Use this formula to find the instantaneous rate of change of \(T\) with respect to \(x\) when \(x=9\) meters.

Step-by-Step Solution

Answer

The instantaneous rate of change of \(T\) at \(x = 9\) is 0.0755.

Step 1: Identify the Problem Type

We need to find the instantaneous rate of change of time \(T\) with respect to height \(x\). This requires finding the derivative of the function \(T=x=0.453 \sqrt{x}\) with respect to \(x\).

Step 2: Express the Function

The function given is \(T = 0.453 \sqrt{x}\). To differentiate this function, it might be helpful to rewrite the square root as a power: \(T = 0.453 x^{1/2}\).

Step 3: Differentiate the Function

Apply the power rule to differentiate \(T = 0.453 x^{1/2}\) with respect to \(x\). The power rule tells us that \(\frac{d}{dx}[x^n] = nx^{n-1}\). Compute the derivative: \[ \frac{dT}{dx} = 0.453 \cdot \frac{1}{2} x^{-1/2} = 0.2265 x^{-1/2}. \]

Step 4: Evaluate the Derivative at x = 9

Substitute \(x = 9\) into the derivative \(\frac{dT}{dx} = 0.2265 x^{-1/2}\) to find the instantaneous rate of change of \(T\) at this point:\[ \frac{dT}{dx}\bigg|_{x=9} = 0.2265 \cdot 9^{-1/2}. \]

Step 5: Simplify the Calculation

Calculate \( 9^{-1/2} = \frac{1}{3} \) since \(9 = 3^2\). Substitute this into the expression to find the derivative:\[ \frac{dT}{dx}\bigg|_{x=9} = 0.2265 \times \frac{1}{3} = 0.0755. \]

Key Concepts

Derivative

A derivative represents the rate at which a function changes as its input changes. In practical terms, it helps us understand how one quantity reacts to changes in another. Take the example of our stuntman where the time to fall, denoted by function \( T \), changes with height \( x \). By finding the rate of change here, we determine how quickly time changes as he falls greater heights.The notation \( \frac{dT}{dx} \) represents the derivative of time with respect to height. This notation indicates the small change in time per unit of change in height. In our scenario, the function \( T = 0.453 \sqrt{x} \) illustrates how time \( T \) grows with the square root of height. By differentiating this function, we can predict the instantaneous rate at which time changes per meter increase in height.

Power Rule

The power rule is a basic tool in calculus for finding derivatives of functions with a certain structure. Specifically, the power rule applies when you have functions in the form \( x^n \). According to the rule, the derivative of \( x^n \) is given by \( nx^{n-1} \). To see it in action, let’s apply it to our stuntman’s function: \( T = 0.453 x^{1/2} \). The function can be differentiated easily by multiplying the exponent \( \frac{1}{2} \) by the constant \( 0.453 \) and then reducing the exponent by one.Applying the power rule here looks like:

• Multiply: \( 0.453 \times \frac{1}{2} = 0.2265 \)
• New exponent: \( x^{1/2 - 1} = x^{-1/2} \)

So, the derivative becomes \( 0.2265 x^{-1/2} \). The power rule simplifies the process of differentiation, making it manageable for functions involving powers of \( x \).

Differentiation

Differentiation is the process of finding a derivative, or the rate at which a function is changing at any given point. It is a fundamental method in calculus used to determine rates of change.In our example with the stuntman, we started with a function describing a scenario: \( T = 0.453 \sqrt{x} \). Differentiation allowed us to find \( \frac{dT}{dx} = 0.2265 x^{-1/2} \), the function that measures how fast time changes as height changes. Once we have the derivative, we can evaluate it at specific points, like when \( x = 9 \) meters, to obtain the instantaneous rate of change at that position:1. Substitute: \( \frac{dT}{dx} \bigg|_{x=9} = 0.2265 \cdot 9^{-1/2} \)2. Simplify: \( 9^{-1/2} = \frac{1}{3} \) because 9 is \( 3^2 \)3. Result: \( \frac{dT}{dx} \bigg|_{x=9} = 0.0755 \)Differentiation thus provides the tools to assess immediate changes accurately, making it invaluable in scientific and engineering calculations.