Problem 88 - Calculus and its applications

Question

Use the Tangent feature from the DRAW menu to find the rate of change in part (b). Pelican Fabrics purchases a new video surveillance system. The value of the system is modeled by $$ V(t)=17,500(0.92)^{t} $$ where \(V\) is the value of the system, in dollars, \(t\) years after its purchase. a) Use the model to estimate the value of the system \(5 \mathrm{yr}\) after it was purchased. b) What is the rate of change in the value of the system at the end of 5 yr? c) When will the system be worth half of its original value?

Step-by-Step Solution

Answer

(a) Approximately $11,615.71. (b) Rate of change is approximately $-968.12 per year at 5 years. (c) Systems value halves around 8.17 years after purchase.

Step 1: Estimate Value After 5 Years

To find the value of the system 5 years after purchase, substitute \( t = 5 \) into the model. Calculate \( V(5) = 17,500 \times (0.92)^{5} \). By solving the equation, \( V(5) \approx 17,500 \times 0.6634 \). Thus, \( V(5) \approx 11,615.71 \).

Step 2: Differentiate to Find Rate of Change

To find the rate of change at \( t = 5 \), differentiate the function \( V(t) = 17,500 \times (0.92)^{t} \) with respect to \( t \). Use the chain rule: \( \frac{dV}{dt} = 17,500 \times \ln(0.92) \times (0.92)^{t} \). This simplifies to \( \frac{dV}{dt} = 17,500 \times (-0.0834) \times (0.92)^{t} \).

Step 3: Calculate Rate of Change at 5 Years

Substitute \( t = 5 \) into the derivative to find the rate of change: \( \frac{dV}{dt} \bigg|_{t=5} = 17,500 \times (-0.0834) \times (0.92)^{5} \). Upon calculation, this results in \( \frac{dV}{dt} \bigg|_{t=5} \approx -968.12 \).

Step 4: Determine When Value is Half

Solve the equation \( V(t) = 17,500 \times (0.92)^{t} = \frac{17,500}{2} \). This simplifies to \( (0.92)^{t} = 0.5 \). Take the natural logarithm of both sides: \( t \ln(0.92) = \ln(0.5) \). Solve for \( t \) as \( t = \frac{\ln(0.5)}{\ln(0.92)} \). Calculating this gives \( t \approx 8.17 \).

Key Concepts

Tangent Line

Imagine a rollercoaster ride, where the track winds up and down. A tangent line at any point on the track gives us a snapshot of the track's exact direction at that point. In calculus, the tangent line assists in understanding geometric concepts on graphs. It acts as the best linear approximation of a curve at a given spot.
A key feature of the tangent line is its slope, which tells us about the rate of change of the curve at that specific point. In simple words:

• The slope of the tangent line shows how steep the curve is at that point.
• If the slope is positive, the curve is climbing up, and if it's negative, the curve is going down.

In our exercise, when evaluating the video surveillance system's value over time, the tangent line gives an instant rate of change of its value at any given year. This hugely simplifies understanding how quickly things are moving at just one particular moment on a complex graph.

Rate of Change

The term 'rate of change' refers to how one quantity changes relative to another. In our daily lives, we often think of speed as a common rate of change - the distance traveled over time. In calculus, we frequently deal with changing quantities and steps:

• The rate of change is reflected by the slope of the tangent line.
• It tells us how fast something is increasing or decreasing at any given point.

In the context of the exercise, the model used to describe the value of the surveillance system over time allows us to determine the rate at which the system loses its value. At the end of five years, using calculus, we calculated the rate of change to be approximately -968.12. This negative value indicates that the system is losing its value at a steady pace of about $968 per year at that time. Visually, you can think of this as the curve of the value graph heading downwards.

Differentiation

Differentiation is the process in calculus used to find the rate at which a function is changing at any point. It plays a central role when we are trying to understand relationships where variables change continuously.
Key points about differentiation:

• Differentiation provides the tool to find the derivative, which is the function that specifically measures the rate of change at any given instance.
• The derivative of a function is essentially the slope of the tangent line to the function's curve.

In the exercise, differentiation performed on the function that models the system's value helps us break down the changes year by year. By using the chain rule on the function \( V(t) = 17,500(0.92)^{t} \), we realized how the value adjusts, giving us the function \( \frac{dV}{dt} = 17,500 \times \ln(0.92) \times (0.92)^{t} \). Substituting \( t = 5 \) indicates how this value decreases, leading us to understand the financial implications and depreciation over time clearly.