Question

A particle starts at \(x=0\) and moves along the \(x\) -axis with velocity \(v(t)=t^{2}+1\) for time \(t \geq 0 .\) Where is the particle at \(t=4 ?\) Approximate the area under the curve using four rectangles of equal width and heights determined by the midpoints of the intervals, as in Example \(1 .\)

Short answer

The particle is at \(x = 76/3\) at \(t = 4\). The approximate area under the curve using four rectangles of equal width and heights determined by the midpoints of the intervals is 10.75.

Step-by-step solution

Calculate Displacement

The displacement of the particle can be obtained from velocity using the concept of integration. In simple terms, the displacement is the area under the velocity-time curve from \(t = 0\) to \(t = 4\). And the velocity-time function given by \(v(t) = t^2 + 1\). Therefore, to find the displacement, calculate \(\int_{0}^{4} v(t) dt =\int_{0}^{4} (t^2 + 1) dt\).

Solve the Integral

The integral can be solved by using the formula for power rule which states \(\int t^n dt = t^{n+1}/(n+1)\). Applying this to the integral, we get \((1/3)*t^3 + t)\)from \(0\) to \(4\) which is \(64/3 + 4 - 0 = 76/3 \). Thus, the particle is at \(x = 76/3\) at \(t = 4\).

Approximate area under the curve

The second part of the problem requires us to approximate the area under the curve using four rectangles of equal width and heights determined by midpoints of the intervals. Divide the interval from \(t = 0\) to \(t = 4\) into 4 segments each of width 1. The midpoints of these intervals are 0.5, 1.5, 2.5 and 3.5. The height of each rectangle is then \(v(t)\) at these points. Thus, the total area, \(A\) can be approximated as \(A = (0.5*(0.5^2)+1) + (1*(1.5^2+1)) + (1*(2.5^2+1)) + (0.5*(3.5^2+1)) = 10.75\). Note that the width and height of the rectangles is different for the first and the last rectangle.

Key concepts

Displacement from Velocity

Calculating the displacement from velocity in calculus is like piecing together a journey from the speed at each point along the way. Displacement represents the change in position of a particle, and it can be found by integrating the velocity function over time.

For instance, if a particle starts at rest and its velocity over time is given by a function like the one provided, you'd want to look at the area under the velocity vs. time graph from the starting time to the final time. In the exercise example, the time interval is between 0 and 4 seconds and the velocity is given by the function v(t) = t^2 + 1. To find the displacement, you calculate the integral of this function from 0 to 4, which gives you the area under the curve—the total distance the particle has traveled in that time frame.

Definite Integral

When we talk about definite integrals, we’re speaking the language of total accumulation. In calculus, a definite integral is used when we want to calculate the total accumulation of a quantity, like displacement. It’s like adding up tiny slices of something over an interval to get the whole picture.

In our exercise, we're interested in the 'whole picture' of displacement from t=0 to t=4. The math magic here is captured by the definite integral, \[\int_{0}^{4} v(t) dt\], which neatly sums up all the instantaneous velocities between these two points in time. This integration process gives you a single value, representing the overall change in position — the sum total of those tiny slices of the journey mentioned before.

Area Under Curve Approximation

Sometimes, calculating the exact area under a curve isn't straightforward, and that's where approximation methods like Riemann sums come into play. Think of it as sketching the shape of a mountain instead of measuring every inch. To approximate the area under our velocity curve, we divide the time interval into smaller sections and estimate the area of each.

In the exercise, we're asked to use the midpoint method to estimate the area under the curve, which means we take the middle value of each time interval to calculate the height of rectangles that represent slices of the area. Adding up the areas of these rectangles gives an estimation of the total area, hence the total displacement. This method isn't perfect—it’s just a sketch, after all—but it provides a good enough idea, especially when dealing with complex functions or when you're without a calculator or computer.

Riemann Sums

Riemann sums are like step ladders that let us climb up to the exact area under a curve. They break down a continuous function into a set of rectangles and the total area of these rectangles is an approximation of the true area. It's a fundamental part of numerical integration and can be likened to counting coins individually to get a total sum.

In the problem we are working on, the approximation of the area under the curve using midpoint Riemann sums is just a methodical way of summing up those 'coins' to estimate how much 'change' in displacement there is over the interval from t=0 to t=4. We find the midpoints, calculate the area of each rectangle, and then total them up to get our approximation of the total displacement.

Power Rule for Integration

The power rule for integration is like a favorite recipe for mathematicians—it simplifies the process of finding the integral of polynomial functions. It's the go-to technique that says if you're integrating t^n, then the antiderivative is t^(n+1)/(n+1).

In our velocity function, we have t^2, and by applying the power rule, we end up with t^(2+1)/(2+1), which simplifies to t^3/3. This is how we bake our polynomial integral cake, so to speak. By following the recipe, we can easily find the area under the curve not only for squared terms but for any power of t.