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Practice Problems for 4/17

1. Let \(f(x)=2+\sin x\). Suppose we approximate the area under \(f(x)\) on \([0,2\pi]\) using right endpoints of the subintervals. On which intervals would the corresponding rectangles give an over-approximation and on which intervals would the area of the rectangles give an under-approximation?

The right endpoints give an over approximation when \(f(x)\) is increasing since the right most point gives the largest \(f\) value in the subinterval. Similarly, right endpoints give an under-approximation when \(f(x)\) is decreasing since the right most point gives the smallest \(f\) value in the subinterval. Therefore, this will give and over-approximation on the interval \([0,\pi/2]\cup[3\pi/2, 2\pi]\) and an under-approximation on \([\pi/2,3\pi/2]\).

2. Let \(v(t)\) be the velocity of a projectile. What does the area under the function \(v(t)\) correspond to, and what does the area under \(|v(t)|\) correspond to?
If \(v(t)\) is ever negative, the area under \(v(t)\) will be negative on some intervals, which takes into account the projectile is moving in a downward direction. Therefore, the area under \(v(t)\) gives the net change in height of the projectile, while the area under \(|v(t)|\) gives the total distance traveled by the projectile, as direction is not taken into account.

3. Find an expression approximating the area of a circle of radius \(r\) using an \(n\) sided polygon, with sides of equal length.

If an \(n\) sided polygon with equal sides is inscribed in a circle, the polygon can be divided up into \(n\) triangles by drawing a line from the center to each vertex of the polygon. The triangles will be identical, and importantly, the angle of the triangle nearest to the center of the circle will be \(2 \pi/n\), as shown in the figure.

The area of the circle is approximately \(n\) times the area of this triangle. To find the area, divide the triangle into two congruent triangles by drawing a perpendicular line which bisects the angle \(2 \pi/n\) as shown in the figure below. Now, the usual trigonometry methods can be used to find the lengths of the side of the triangle, and thus the area of the triangle.

Each of the smaller triangles has area \(\frac{1}{2}bh=\frac{1}{2}r\sin(\pi/n)r\cos(\pi/n)\) and the area of the circle is approximately \(nr^2\sin(\pi/n)\cos(\pi/n)\).