Problem 82

Question

Solve the initial value problems in Exercises $67-86$. $$ \frac{d^{2} s}{d t^{2}}=\frac{3 t}{8} ;\left.\quad \frac{d s}{d t}\right|_{t=4}=3, \quad s(4)=4 $$

Step-by-Step Solution

Verified

Answer

The solution is $ s(t) = \frac{t^3}{16} $.

1Step 1: Integrate the Second Derivative

The initial differential equation is $ \frac{d^2 s}{dt^2} = \frac{3t}{8} $. Integrate this equation with respect to $ t $ to find the first derivative, $ \frac{ds}{dt} $.\[ \frac{ds}{dt} = \int \frac{3t}{8} \, dt = \frac{3}{8} \int t \, dt = \frac{3}{8} \left( \frac{t^2}{2} \right) + C_1 \]Simplifying, we get:\[ \frac{ds}{dt} = \frac{3t^2}{16} + C_1 \]

2Step 2: Apply the First Initial Condition

Given $ \left. \frac{ds}{dt} \right|_{t=4} = 3 $, substitute $ t = 4 $ into the equation found from integration:\[ 3 = \frac{3 \times 4^2}{16} + C_1 \]Simplifying gives:\[ 3 = \frac{48}{16} + C_1 \]\[ 3 = 3 + C_1 \]Therefore, $ C_1 = 0 $. So the equation becomes:\[ \frac{ds}{dt} = \frac{3t^2}{16} \]

3Step 3: Integrate Again for the Function

Now integrate $ \frac{ds}{dt} = \frac{3t^2}{16} $ to find $ s(t) $:\[ s(t) = \int \frac{3t^2}{16} \, dt = \frac{3}{16} \int t^2 \, dt = \frac{3}{16} \left( \frac{t^3}{3} \right) + C_2 \]Simplifying, we get:\[ s(t) = \frac{t^3}{16} + C_2 \]

4Step 4: Apply the Second Initial Condition

Using the condition $ s(4) = 4 $, substitute $ t = 4 $ into the equation for $ s(t) $:\[ 4 = \frac{4^3}{16} + C_2 \]This simplifies to:\[ 4 = \frac{64}{16} + C_2 \]\[ 4 = 4 + C_2 \]Thus, $ C_2 = 0 $. Therefore, the function is:\[ s(t) = \frac{t^3}{16} \]

Key Concepts

Differential EquationIntegrationInitial ConditionsSecond Derivative

Differential Equation

A differential equation involves derivatives of a function and describes how a particular quantity changes with respect to another. In our exercise, the differential equation is given as $ \frac{d^2 s}{dt^2} = \frac{3t}{8} $. This specific form is a second-order differential equation, indicating that it contains the second derivative $ \frac{d^2 s}{dt^2} $ of a function $ s(t) $ with respect to the variable $ t $.

The "second-order" term denotes that it relates to a rate of change of a rate of change, essentially how acceleration or curvature behaves in a mathematical context.
The function $ s(t) $ we are seeking represents some quantity evolving over time, based on the information provided by the differential equation.

Differential equations are fundamental in describing natural phenomena, from physics to engineering, and they are essential for forming and solving models of real-world problems.

Integration

Integration is the process used to reverse differentiation, and it helps in finding a function when its derivative is known. In our step-by-step solution, integration was used twice.

The first integration was performed on the second derivative $ \frac{d^2 s}{dt^2} $ to get the first derivative $ \frac{ds}{dt} $.
The result was $ \frac{ds}{dt} = \frac{3t^2}{16} + C_1 $, where $ C_1 $ is an integration constant. Constants emerge during integration because derivatives "lose" constant terms.
The second integration was carried out on $ \frac{ds}{dt} $ to obtain $ s(t) $, resulting in $ s(t) = \frac{t^3}{16} + C_2 $, again with $ C_2 $ as an integration constant.

Integration allowed us to move stepwise from the second derivative back to the original function, faithfully reconstructing the function $ s(t) $ from its given rates of change.

Initial Conditions

Initial conditions provide specific values that allow us to solve for the integration constants that appear during the process of integration. These conditions anchor the solution to a known part of the function's behavior.

In this problem, the first initial condition had us evaluate $ \frac{ds}{dt} $ at $ t = 4 $ where it equaled 3. Using this, we found $ C_1 = 0 $.
The second initial condition, $ s(4) = 4 $, was used to determine $ C_2 $ after integrating again, which also turned out to be 0.

By applying these initial conditions, we effectively "solved" the functions for their specific constants, giving us a precise expression of $ s(t) $ that fits the particular set of criteria provided.

Second Derivative

The second derivative in a differential equation represents the rate of change of the first derivative, effectively describing how a function's rate of change itself changes over time.

In physical terms, if $ s(t) $ were a position function, $ \frac{d^2 s}{dt^2} $ would describe acceleration, or how the velocity is changing at any point in time.
In the given equation $ \frac{d^2 s}{dt^2} = \frac{3t}{8} $, this suggests that the function's acceleration at any time $ t $ is linearly dependent on time itself.
Understanding this concept is crucial, as it forms the foundational step from which we derive $ \frac{ds}{dt} $ and subsequently $ s(t) $ through integration, ultimately solving the initial value problem.

Thus, the second derivative sets the stage for understanding future behavior of the function in terms of both velocity and position.

Problem 81

Problem 82

Other exercises in this chapter

Problem 81

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Problem 81

Parabolas a. Find the coordinates of the vertex of the parabola $\quad y=a x^{2}+b x+c, a \neq 0$ b. When is the parabola concave up? Concave down? Give reaso

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Problem 82

Is it true that the concavity of the graph of a twice-differentiable function $y=f(x)$ changes every time $f^{\prime \prime}(x)=0 ?$ Give reasons for your a

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Problem 83

Solve the initial value problems in Exercises $67-86$. $$ \frac{d^{3} y}{d x^{3}}=6 ; \quad y^{\prime \prime}(0)=-8, \quad y^{\prime}(0)=0, \quad y(0)=5 $$

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