91Ó°ÊÓ

Understanding how to verify if a function is a solution to a given differential equation is crucial in mathematics. Let's break down part (a) of the exercise.
We have the differential equation \( \frac{d y}{d t}=y-t \).

Consider the following functions to check if they are solutions to this differential equation:

• \text{ - } y(t)=t+1+2e^{t}
• \text{ - } y(t)=t+1
• \text{ - } y(t)=t+2

To verify, we need to:

• Find the derivative \( \frac{d y}{d t} \) of each function
• Substitute both \( y \) and \( \frac{d y}{d t} \) into the given differential equation
• Check if the equation holds true

For the first function:

• \( y(t) = t + 1 + 2e^t \)
• \( \frac{d y}{d t} = 1 + 2e^t \)
• Substitute to get \( 1 + 2e^t = (t + 1 + 2e^t) - t \)
• It simplifies to \( 1 + 2e^t = 1 + 2e^t \)

So \( y(t) = t + 1 + 2e^t \) is a solution.
For the second function:

• \( y(t) = t + 1 \)
• \( \frac{d y}{d t} = 1 \)
• Substitute to get \( 1 = (t + 1) - t \)
• It simplifies to \( 1 = 1 \)

Thus, \( y(t) = t + 1 \) is a solution.
For the third function:

• \( y(t) = t + 2 \)
• \( \frac{d y}{d t} = 1 \)
• Substitute to get \( 1 = (t + 2) - t \)
• It simplifies to \( 1 = 2 \)

Therefore, \( y(t) = t + 2 \) is not a solution.

By following these clear steps, one can determine whether a function is a solution to a given differential equation.

The spring constant, denoted by \( k \), plays a significant role in the behavior of springs and oscillations. It is a measure of the stiffness of the spring in the scale. Let's consider part (b) of the exercise.

The height of the bananas, given by \( h \), in the scale is described by the differential equation:
\(\frac{d^{2}h}{d t^{2}}=-k h\)

Given the function \( h(t) = 4 \text{sin}(3t) \), we need to find \( k \).

Steps to find the spring constant:

• Compute the first derivative \( \frac{d h}{d t} \)
• Compute the second derivative \( \frac{d^{2}h}{d t^{2}}\)
• Substitute the second derivative into the differential equation and solve for \( k \)

First derivative:
\( \frac{d h}{d t} = 12 \text{cos}(3t) \)
Second derivative:
\( \frac{d^{2}h}{d t^{2}} = -36 \text{sin}(3t) \)
Substitute into the differential equation:
\(-36 \text{sin}(3t) = -k \text{4 sin}(3t)\)
Simplify to find \( k \):
\( k = \frac{36}{4} = 9 \)

Therefore, the spring constant of the scale is \( k = 9 \). Understanding this concept helps in analyzing the properties of springs and the forces involved in oscillating systems.

Derivatives are a fundamental concept in calculus, representing rates of change. Let's explore their role in solving differential equations, as seen in the exercise.

A differential equation involves derivatives which describe how a function changes. In part (a), we used first derivatives, while in part (b), second derivatives were also involved.

First, let's understand a few key points:

• First Derivative: Denoted as \( \frac{d y}{d t} \), it shows the rate of change of \( y \) with respect to \( t \).
• Second Derivative: Denoted as \( \frac{d^{2} h}{d t^{2}} \), it shows the rate of change of the first derivative, hence describing the acceleration or curvature of the original function.

For example, in part (b), we calculated both derivatives for the function:
\( h(t) = 4 \text{sin}(3t) \)

The steps involved were:

• First derivative: \( \frac{d h}{d t} = 12 \text{cos}(3t) \)
• Second derivative: \( \frac{d^{2}h}{d t^{2}} = -36 \text{sin}(3t) \)

These derivatives were then substituted back into the differential equation to verify the solution or find constants, such as the spring constant \( k \) in part (b).

Therefore, mastering derivatives is crucial to solving differential equations, understanding rates of change, and analyzing the behavior of functions. By practicing these steps, students can strengthen their grasp on how derivatives interact within these equations.