Problem 33 - Calculus Early Transcendentals

Question

Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and initial position. $$a(t)=\cos 2 t, v(0)=5, s(0)=7$$

Step-by-Step Solution

Answer

Answer: The position function is \(s(t) = -\frac{1}{4} \cos{2t} + 5t + \frac{29}{4}\) and the velocity function is \(v(t) = \frac{1}{2} \sin{2t} + 5\).

Step 1: Integrate acceleration to find the velocity function

To find the velocity function, integrate the acceleration function: $$ v(t) = \int a(t) \: dt = \int \cos{2t} \: dt $$ We will need to use integration by substitution, let: $$ u = 2t $$ $$ \frac{du}{dt} = 2 \Rightarrow dt = \frac{1}{2} du $$ Now, substitute and integrate: $$ v(t) = \int \cos{u} \cdot \frac{1}{2} \: du = \frac{1}{2} \int \cos{u} \: du = \frac{1}{2} \sin{u} + C $$ Now, substitute back \(u = 2t\): $$ v(t) = \frac{1}{2} \sin{2t} + C $$

Step 2: Apply the initial condition for velocity

We know that \(v(0) = 5\). Substitute \(t=0\) into our velocity function and solve for C: $$ 5 = \frac{1}{2} \sin{2(0)} + C $$ $$ 5 = 0 + C $$ $$ C = 5 $$ Now, we have the velocity function: $$ v(t) = \frac{1}{2} \sin{2t} + 5 $$

Step 3: Integrate velocity to find the position function

To find the position function, integrate the velocity function: $$ s(t) = \int v(t) \: dt = \int \left ( \frac{1}{2} \sin{2t} + 5 \right ) \: dt $$ Now, integrate term by term: $$ s(t) = \frac{1}{2} \int \sin{2t} \: dt + \int 5 \: dt $$ Again, use integration by substitution for the first term, already done in Step 1: $$ s(t) = \frac{1}{2} \cdot \left ( -\frac{1}{2} \cos{2t} \right ) + 5t + D $$ Simplify the expression: $$ s(t) = -\frac{1}{4} \cos{2t} + 5t + D $$

Step 4: Apply the initial condition for position

We know that \(s(0) = 7\). Substitute \(t=0\) into our position function and solve for D: $$ 7 = -\frac{1}{4} \cos{2(0)} + 5(0) + D $$ $$ 7 = -\frac{1}{4} + D $$ $$ D = \frac{29}{4} $$ Now, we have the position function: $$ s(t) = -\frac{1}{4} \cos{2t} + 5t + \frac{29}{4} $$ The position function is \(s(t) = -\frac{1}{4} \cos{2t} + 5t + \frac{29}{4}\), and the velocity function is \(v(t) = \frac{1}{2} \sin{2t} + 5\).

Key Concepts

Integration by Substitution

Integration by substitution is a method used to simplify complex integrals. It's like the mathematical equivalent of swapping out hard-to-peel vegetables with easier ones before you start cooking. In calculus, this technique allows us to change the variable of integration to something friendlier, which often results in an integral we can solve more straightforwardly.

Let's take a closer look at the exercise where we used integration by substitution to find the velocity function from the acceleration. Given the acceleration function as \( a(t)=\cos 2 t \), and needing to find the integral of \( \cos{2t} \), the process begins by setting \( u=2t \). This small shift transforms the problem into something much easier to work with, essentially peeling back the layers of the original, more complex equation.

After substituting \( dt \), with \( \frac{1}{2} du \), integrating becomes a breeze - just a simple integral of \( \cos{u} \). You end up with \( \frac{1}{2} \sin{u} \), which is far easier to manage than our starting equation. This example beautifully illustrates how integration by substitution is a powerful tool for making complex calculus problems much more approachable.

Initial Value Problems

Initial value problems are puzzles in calculus that come with a piece of the solution: an initial condition. It's like being given a snapshot of a moving train at a certain time, and your job is to figure out the train's entire journey. In the context of motion along a straight line, these initial conditions can be things like an object's initial position or its initial velocity at time \( t=0 \).

In our exercise, we had the initial conditions of velocity, \( v(0)=5 \), and position, \( s(0)=7 \). These were the 'snapshots' that helped us solve the rest of the problem. We used the initial velocity to find the constant of integration for the velocity function. Then, we similarly used the initial position to determine the constant for the position function. With the initial conditions laid out, it becomes a matter of simply plugging in values and solving algebraic equations to get those constants, \( C \), and \( D \) which untangle the functions we're looking to define.

Velocity and Position Function

In physics, understanding how objects move is all about the velocity and position functions. They're the bread and butter of kinematics - the study of motion without considering its causes. In calculus terms, velocity is the derivative of the position function, and position is the integral of the velocity function.

Returning to our textbook exercise, we integrated the given acceleration to find the velocity function - a method akin to retracing steps back to see how fast we were going after accelerating at a known rate. Next, we integrated our newfound velocity function to discover the object's position. You could see this as compiling a travelogue from a list of speeds, documenting where our object is at any given time.

This integral dance between velocity and position functions gives us a full picture of motion along a straight line. By understanding these concepts deeply, we can predict the future location of moving objects, or we could even rewind time to figure out where they started. It's calculus giving us a superpower to navigate through the dimensions of motion!