Problem 27 - Calculus

Question

True or false. If false, spy what's true. (a) The time for \(y=e^{e t}\) to double is \((\ln 2) /(\ln c)\). (b) If \(y^{\prime}=c y\) and \(z^{\prime}=c z\) then \(\left.(y+z)^{\prime}=2 d y+z\right)\) (c) If \(y^{\prime}=c y\) and \(z^{\prime}=c z\) then \((y / z)^{\prime}=0\) (d) If \(y^{\prime}=c y\) and \(z^{\prime}=C z\) then \((y z)^{\prime}=(c+C) y z\)

Step-by-Step Solution

Answer

(a) False: doubling time is \(\ln 2/e\). (b) False. (c) True. (d) True.

Step 1: Understand part (a)

We need to determine if the given formula \(t_\text{double} = \frac{\ln 2}{\ln c}\) is the correct time for the function \(y = e^{et}\) to double. To find the doubling time, we usually set \(y(t+T) = 2y(t)\). However, for the exponential function \(y = e^{e t}\), the doubling time is calculated differently. The correct formula is \(T = \frac{\ln 2}{e}\). The provided formula \(\frac{\ln 2}{\ln c}\) applies to exponential growth with base \(c\), not \(e^e\). Thus, this assertion is false.

Step 2: Correct part (a)

The true statement, given that part (a) is false, is related to the doubling time for a function of the form \(y = c^t\). The doubling time in that scenario is \(t_\text{double} = \frac{\ln 2}{\ln c}\).

Step 3: Verify part (b)

The statement claims \((y+z)' = 2dy + z\). When given \(y' = cy\) and \(z' = cz\), using the linearity of derivatives, \((y+z)' = y' + z'\). Therefore, \((y+z)' = cy + cz\), not \(2dy + z\). The statement is false.

Step 4: Understand part (c)

We are evaluating if \((y/z)' = 0\). By quotient rule \((y/z)' = (y'z - yz')/z^2\). Given \(y' = cy\) and \(z' = cz\), we substitute and simplify: \((y'z - yz') = (cyz - czy) = 0\). Therefore, \((y/z)' = 0\) is true.

Step 5: Verify part (d)

Evaluate if \((yz)' = (c+C)yz\) is true. Using the product rule, \((yz)' = y'z + yz' = cyz + Czy = (c+C)yz\). This confirms that the statement is true.

Key Concepts

Exponential Growth

Exponential growth is a key concept in differential equations, often observed in processes like population dynamics or finance. It occurs when the rate of change of a quantity is proportional to the current amount of that quantity. This is typically expressed through the equation \( y = a e^{bx} \) where \( a \) and \( b \) are constants. The nature of exponential growth is that it increases continually and rapidly, doubling over regular intervals known as doubling time. Understanding this concept is crucial for solving problems involving exponential functions. Here, the time it takes for a function to double can be misleading, as seen in the function \( y = e^{et} \), where normal rules of exponential doubling don't apply, requiring special consideration of the constant base used.

Product Rule

The product rule is a fundamental technique in calculus used to differentiate expressions where two functions are multiplied together. If we have two differentiable functions \( u(x) \) and \( v(x) \), the derivative of their product is given by: \( (uv)' = u'v + uv' \). This rule comes into play frequently in problems involving rates of change and exponential functions. In the original problem's context, it helps confirm the accuracy of derivative expressions like \( (yz)' = (c+C)yz \) by ensuring that each part of the product is differentiated correctly. The product rule thus ensures that we can correctly compute changes in combinations of variables, maintaining the integrity of the calculus methods.

Quotient Rule

The quotient rule is essential while working with division in calculus, particularly in differential equations. It aids in differentiating ratios of functions. Given two differentiable functions \( u(x) \) and \( v(x) \), where \( v(x) eq 0 \), the quotient rule is given by: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). This formula highlights how the derivative of a division differs from mere subtraction. In our original exercise, it allowed us to assess that under certain conditions, namely when the ratios of derivatives conform, the derivative of the ratio can be zero, simplifying the problem's complexity. This utility of the quotient rule in simplifying differential equations cannot be overstated.

Doubling Time

Doubling time is a fascinating concept linked closely with exponential growth. It refers to the period required for a quantity undergoing exponential growth to double its size. This is calculated using the formula \( T = \frac{\ln 2}{r} \), where \( r \) is the growth rate. In practical scenarios, doubling time helps us understand the dynamics of rapid processes, from financial investments to biological growth. In the provided exercise, the misunderstanding in part (a) about the base of the exponential function highlights the nuances in calculating doubling time. While the usual formula applies to functions like \( y = c^t \), with an inherent growth rate, problems like \( y = e^{et} \) require a tailored approach, illustrating the importance of the base in these calculations.