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A $$\$ 10,000$$ automobile depreciates so that its value after \(t\) years is \(V(t)=10,000 e^{-0.35 t}\) dollars. Find the instantaneous rate of change of its value: a. when it is new \((t=0)\). b. after 2 years.

Short Answer

Expert verified
a. \(-3500\) dollars/year b. \(-1738.1\) dollars/year

Step by step solution

01

Understand the problem

The problem asks for the instantaneous rate of change, which implies finding the derivative of the given function. The function is \(V(t) = 10,000 e^{-0.35t}\). We will find \(V'(t)\), the derivative with respect to \(t\), to solve this problem.
02

Differentiate the Function

To find \(V'(t)\), differentiate \(V(t) = 10,000 e^{-0.35t}\) using the chain rule. The derivative of \(e^{ax}\) with respect to \(x\) is \(a e^{ax}\). Thus, \(V'(t) = 10,000 \times (-0.35)e^{-0.35t}\). Simplifying gives \(V'(t) = -3500e^{-0.35t}\).
03

Calculate the instantaneous rate when \(t=0\)

Substitute \(t = 0\) into the derivative \(V'(t) = -3500e^{-0.35t}\). This gives \(V'(0) = -3500e^{-0.35(0)} = -3500e^{0} = -3500 \). This means the instantaneous rate of change when \(t = 0\) is \(-3500\) dollars per year.
04

Calculate the instantaneous rate when \(t=2\)

Substitute \(t = 2\) into the derivative \(V'(t) = -3500e^{-0.35t}\). This gives \(V'(2) = -3500e^{-0.35(2)} = -3500e^{-0.7}\). Using a calculator, \(e^{-0.7} \approx 0.4966\), so \(V'(2) \approx -3500 \times 0.4966 = -1738.1\). Thus, the instantaneous rate of change after 2 years is approximately \(-1738.1\) dollars per year.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Instantaneous Rate of Change
The instantaneous rate of change is a crucial concept in calculus. It tells us how fast a value is changing at a specific moment. Imagine driving a car. The speedometer shows your speed at a particular instant; this is similar to what the instantaneous rate of change does for functions.
To find this rate, we need to calculate the derivative of the function. The derivative gives us the slope of the tangent line at any point on the function's curve. It tells us how steeply the value is increasing or decreasing.
In the case of the automobile's value function, the instantaneous rate of change at a specific time gives us how fast the car's value is depreciating at that precise moment. By substituting particular values of time into the derivative, we can find exactly how much the value changes per year at those times.
Derivative
The derivative is one of the most important aspects of calculus. It's essentially a tool that allows us to measure change. When we differentiate a function, we are essentially finding a formula that describes how the function's output changes based on its input.
The derivative also helps in understanding the behavior of the graph. If the derivative is positive, the function is increasing; if it's negative, the function is decreasing. For our car's depreciation function, this means at any given year, the derivative tells us how quickly the car's value is dropping.
In mathematical terms, the derivative of a function at a point provides the slope of the tangent to the curve at that point. This slope indicates how sharply the curve is rising or falling there. Calculating the derivative might involve different rules, such as the basic rules of differentiation or more advanced concepts like the chain rule.
Chain Rule
The chain rule is a fundamental technique used in calculus to differentiate compositions of functions. When you have a function within another function, like in the expression for the automobile's value, the chain rule helps us find the derivative efficiently.
For the function \(V(t) = 10,000 e^{-0.35t}\), the chain rule is necessary because there's a function inside another function—the exponential function \(e^{x}\) with a variable exponent \(-0.35t\).
The chain rule states that to differentiate a composite function, you take the derivative of the outer function and multiply it by the derivative of the inner function. So, for differentiating \(e^{-0.35t}\), we first take the derivative of \(e^{x}\) as \(e^{x}\) and multiply it by the derivative of \(-0.35t\), which is \(-0.35\).
This results in \(-0.35e^{-0.35t}\). The overall derivative then becomes \(-3500e^{-0.35t}\) after considering the multiplier \(10,000\), giving the instantaneous rate of change of the car's value over time.

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Most popular questions from this chapter

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