Problem 26 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For what value(s) of \(\beta\), if any, is (a) \(y=C_{1} \sin \beta t\) a solution to \(y^{\prime \prime}=16 y\) ? (b) \(y=C_{2} \cos \beta t\) a solution to \(y^{\prime \prime}=16 y\) ? (c) \(y=C_{3} e^{\beta t}\) a solution to \(y^{\prime \prime}=16 y\) ?

Step-by-Step Solution

Answer

The value of \(\beta\) that makes each of the given functions a solution to \(y'' = 16y\) is \(\pm 4\).

Step 1: Analyze the first function

Let's start with the first function. The second derivative of \(y=C_{1} \sin \beta t\) is found using chain rule, yielding \(y'' = -\beta ^2 C_{1} \sin \beta t\).

Step 2: Compare to original function

Comparing this to 16 times the original function, we have that \(-\beta ^2 C_{1} \sin \beta t = 16 C_{1} \sin \beta t\). Solving for \(\beta ^2\), we find \(\beta = \pm 4\).

Step 3: Analyze the second function

Now, look at the second function. The second derivative of \(y=C_{2} \cos \beta t\) is \(- \beta ^2 C_{2} \cos \beta t \).

Step 4: Compare to original function

Again, comparing this to 16 times the original function, we find that \(\beta ^2 = 16 \), so \(\beta = \pm 4\).

Step 5: Analyze the third function

Finally, consider the third function. The second derivative of \(y=C_{3} e^{\beta t}\) is \(\beta ^2 C_{3} e^{\beta t}$. Therefore, the derivative is equal to 16 times the function when \(\beta^2 = 16 \).

Step 6: Solve for beta

Solving for \(\beta ^2 = 16\) yields \(\beta = \pm 4\).

Key Concepts

Second Derivative

In the realm of differential equations, the second derivative signifies the rate at which the first derivative changes. In simpler terms, it measures how the slope of the function's tangent line is changing. For the given exercise, you start by finding the second derivative of various functions. This requires using foundational calculus tools like the chain rule.

• For a sinusoidal function like \(y = C_1 \sin \beta t\), the second derivative is \(y'' = -\beta^2 C_1 \sin \beta t\). This transformation shows how the wave's angle impacts its behavior.
• For \(y = C_2 \cos \beta t\), similarly, we get \(y'' = -\beta^2 C_2 \cos \beta t\). This calculation underscores the mirroring characteristics of sine and cosine functions' derivatives.
• The exponential function \(y = C_3 e^{\beta t}\) translates into \(y'' = \beta^2 C_3 e^{\beta t}\), displaying how exponential growth or decay rates re-emerge in their higher-order derivatives.

The resulting forms of these second derivatives are pivotal for matching them to the equation \(y'' = 16y\), revealing the influence of the parameter \(\beta\).

Trigonometric Functions

Trigonometric functions like sine and cosine describe oscillations and waves. In calculus, these functions frequently appear due to their periodic nature. Understanding their behavior, especially regarding derivatives, is crucial.

• The sine function, \(y = C_1 \sin \beta t\) moves through positive and negative cycles. By differentiating this function, its derivatives reflect repeated patterns that shift with multiplication by constants like \(-\beta^2\).
• For a cosine function \(y = C_2 \cos \beta t\), the process mirrors that of sine, yielding similar oscillatory derivative forms \(-\beta^2 C_2 \cos \beta t\) due to their inherent relationship \(\cos(x) = \sin(x + \pi/2)\).

The alignment of these derivatives to the form \(y'' = 16 y\) necessitates solving for specific values of \(\beta\) that harmonize their frequencies with the equation's demands. In this case, \(\beta = \pm 4\) achieves this balance, maintaining the periodic nature intrinsic to trigonometric expressions.

Exponential Functions

Exponential functions represent a constant rate of growth or decay, expressed via base \(e\), the natural logarithm's foundation. When dealing with differential equations, exponential functions often highlight the fundamental balance between compounding processes and changes in state.

• The form \(y = C_3 e^{\beta t}\) results in a second derivative of \(y'' = \beta^2 C_3 e^{\beta t}\). Here, the characteristic "\(e\)-to-the-power-of" structure reflects unchanged terms, aside from included scaling factors like \(\beta^2\).
• This special trait of exponential functions - that their derivatives don't alter the base structure - is a major reason they simplify solving homogenous linear differential equations.

In the context of matching an equation like \(y'' = 16y\), you find that choosing \(\beta = \pm 4\) ensures that the exponential's rapid change is consistent with the required conditions. This example illustrates the beauty of exponentials in maintaining form while adjusting growth rates through scaling factors.