Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Show that if \(A, B\) and \(\omega\) are constants and \(y(t)=A \sin (\omega t)+B \cos (\omega t),\) then \(y(0)=B \quad y^{\prime}(0)=\omega A\) and $$ y^{\prime \prime}(t)+\omega^{2} y(t)=0 $$

Step-by-Step Solution

Answer

Yes, both conditions and the differential equation hold true.

Step 1: Evaluate y(0)

Evaluate the function at \(t = 0\). The function is \(y(t) = A \sin(\omega t) + B \cos(\omega t)\). Substitute \(t = 0\) to get \(y(0) = A \sin(0) + B \cos(0)\). Since \(\sin(0) = 0\) and \(\cos(0) = 1\), we find that \(y(0) = B\).

Step 2: Differentiate y(t) to find y'(t)

Differentiate \(y(t) = A \sin(\omega t) + B \cos(\omega t)\) with respect to \(t\). Using the chain rule, the derivative of \(\sin(\omega t)\) is \(\omega \cos(\omega t)\) and the derivative of \(\cos(\omega t)\) is \(-\omega \sin(\omega t)\). Thus, \(y'(t) = A \omega \cos(\omega t) - B \omega \sin(\omega t)\).

Step 3: Evaluate y'(0)

Evaluate the derivative at \(t = 0\). Substitute \(t = 0\) into \(y'(t) = A \omega \cos(\omega t) - B \omega \sin(\omega t)\) to get \(y'(0) = A \omega \cos(0) - B \omega \sin(0)\). \(\cos(0) = 1\) and \(\sin(0) = 0\), so \(y'(0) = \omega A\).

Step 4: Differentiate y'(t) to find y''(t)

Differentiate \(y'(t) = A \omega \cos(\omega t) - B \omega \sin(\omega t)\) to find \(y''(t)\). The derivative of \(\cos(\omega t)\) is \(-\omega \sin(\omega t)\) and the derivative of \(-\sin(\omega t)\) is \(-\omega \cos(\omega t)\). Therefore, \(y''(t) = -A \omega^2 \sin(\omega t) - B \omega^2 \cos(\omega t)\).

Step 5: Verify the given differential equation

Verify that \(y''(t) + \omega^2 y(t) = 0\). We have \(y''(t) = -A \omega^2 \sin(\omega t) - B \omega^2 \cos(\omega t)\) and \(y(t) = A \sin(\omega t) + B \cos(\omega t)\). Substitute these into the equation: \(-A \omega^2 \sin(\omega t) - B \omega^2 \cos(\omega t) + \omega^2 (A \sin(\omega t) + B \cos(\omega t)) = 0\). The terms cancel out, verifying the differential equation.

Key Concepts

Trigonometric Functions

Trigonometric functions are used to relate angles to sides of right triangles and appear frequently in wave functions, like in our exercise where you see sine and cosine.

• Sine Function: Represents the y-coordinate of a unit circle as the angle varies.
• Cosine Function: Represents the x-coordinate of a unit circle with angle variation.

In our given function, \(y(t) = A \sin(\omega t) + B \cos(\omega t)\), \(\omega\) is the angular frequency. It dictates how rapidly the trigonometric functions oscillate through their cycles. When we evaluate \(y(t)\) at \(t = 0\), only the cosine part contributes, as \(\sin(0) = 0\). Therefore, \(y(0) = B\), demonstrating the value of \(y\) at the initial time without any effect from \(A\). Understanding these basics helps in analyzing periodic behavior in physics and engineering.

Chain Rule

The chain rule is a fundamental tool in calculus used for differentiating compositions of functions. In essence, you perform a derivative of the outer function while multiplying it by the derivative of the inner function. In the context of our problem, when differentiating \(y(t) = A \sin(\omega t) + B \cos(\omega t)\), the chain rule guides us through:

• Differentiating \(\sin(\omega t)\): It's outer part is \(\sin\), giving \(\cos(\omega t)\), and multiplying by the derivative of the inner part, which is \(\omega\), results in \(\omega \cos(\omega t)\).
• Differentiating \(\cos(\omega t)\): The outer function \(\cos\) turns into \(-\sin(\omega t)\), multiplied by the derivative of the inner, \(\omega\), results in \(-\omega \sin(\omega t)\).

The product of these calculations provides \(y'(t) = A \omega \cos(\omega t) - B \omega \sin(\omega t)\). Knowing how to apply the chain rule expands problem-solving capabilities across calculus topics.

Initial Conditions

Initial conditions are values specified at the start of a problem to provide a particular solution to differential equations. They decide the particular trajectory that a system follows. In our example:

• For \(y(0)\): We found the trigonometric function's initial value by substituting \(t = 0\). This gave \(y(0) = B\), representing an initial position or state of the system.
• For \(y'(0)\): Differentiating \(y(t)\) produced \(y'(t) = A \omega \cos(\omega t) - B \omega \sin(\omega t)\). Evaluating this at \(t = 0\), the derivative is \(\omega A\), implying an initial rate of change or velocity at time zero.

By providing specific starting values, initial conditions are crucial in determining the exact form of a solution with respect to time or another independent variable.

Second Derivative

The second derivative indicates how the rate of change of a function is itself changing. In physical contexts, it often relates to acceleration or concavity of the curve. For the problem, we need to find \(y''(t)\). Starting with the first derivative \(y'(t) = A \omega \cos(\omega t) - B \omega \sin(\omega t)\), applying derivatives again requires the chain rule:

• The derivative of \(\cos(\omega t)\) is \(-\omega \sin(\omega t)\), leading to \(-A \omega^2 \sin(\omega t)\).
• The derivative of \(-\sin(\omega t)\) is \(-\omega \cos(\omega t)\), contributing \(-B \omega^2 \cos(\omega t)\).

Thus, \(y''(t) = -A \omega^2 \sin(\omega t) - B \omega^2 \cos(\omega t)\). When put back into the differential equation \(y''(t) + \omega^2 y(t) = 0\), the terms cancel, verifying the given equation is a proper motion equation. Comprehension of second derivatives allows us to grasp systems' dynamics and reactions better.