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Applications of integration

Linear Motion

Velocity v(t), and acceleration a(t) are the first and second derivatives of the displacement s(t) of an object moving along a line.

The distance, s, covered by a moving object is the area under the velocity curve, and the final velocity of the object can be found by integrating the acceleration curve.

$$Δs=∫vdt$$ $$Δv_f=∫adt$$

where Δ means 'change in'.

Care must be taken not to confuse distance covered and final displacement from the initial position. The distance covered by a moving particle is the total area under the velocity-time curve. The area under the x-axis (time) is still positive distance travelled. In terms of displacement, the curve under the x-axis represents negative velocity, so the particle has reversed direction, and will reduce the displacement. If the two areas, above and below the x-axis of the velocity-time curve, are equal, the particle finds itself back at the starting position, yet has covered a positive distance in total.

Only if the velocity does not change sign (remains positive or negative for all Δt (change in time)), will the displacement equal the distance travelled. If the velocity changes sign, it has started to move in the opposite direction, so it displacement changes sign as well.

The same is true for speed and velocity. Speed and distance travelled are scalar value, so have no sign. Velocity and displacement are vectors, so have a sign in linear motion, and direction in 2D and 3D space.

Minimum distance from a point to a curve

To find the minimum distance between a point $P(x_p, y_p)$ and a point C on a curve $y = f(x)$, use Pythagoras: $PC = √{(x_c-x_p)^2 + (y_c-y_p)^2}$.

Now the clever bit: if you square both sides of the equation, it will not change the minimum distance values.

Find the differential of $PC^2 = (x_c-x_p)^2 + (y_c-y_p)^2$, and set it to zero, to find the minimum, which provides the x value of the point which is closest to the curve.

Find the minimum distance between the point P(2,4) and the curve $y = 3sin(3x)$.

Let us find a function to describe the distance between the point and the curve $y = 3sin(3x)$:

$d^2(d) = (2-x)^2+(4-3sin(3x))^2$

Notice we are actually using the square of the distance: this doesn't affect the result but simplifies calculations greatly. Next, we equate the first derivative to zero to find the minimum value of $d^2(x)$:

$f'(d^2(x)) = -2(2-x)-18cos(3x)(4-3sin(3x)) = 0$

Maximum and minimum of a function
Minimum distance from a point to a curve

By solving this equation, we find that the shortest distance is at $x=2.6$ and has a value of $√{d^2(2.6)} ≅ 1.17$.

Minima and Maxima

The greatest area of a rectangle under a curve can be found by differentiating the equation for the rectangle in terms of the variable in the curve (usually x), and setting the differential to zero to find the maxima.

This is typical of periodic motion, such as sine and cosine, which represent rotations. A point on the circumference of a bobbin returns with each rotation to the same position, but the length of twine it is winding continues to increase linearly.

The area under the sine curve: $A = ∫_0^{2π} |sinx|dx$. Without the absolute value signs, the total area would be zero.

Optimisation of area F(x) = xcos(2x).

Volume of Revolution

A solid is formed by rotating an area between a curve and the $x$-axis through 360° (2π radians):

$$V=π∫_a^by^2dx$$

The volume of the solid formed by the rotation of $y=cosx$, between $x=0$ and $x={π}/2$ around the x-axis is:

$V= π∫_0^{π/2}cos^2xdx = π[1/2(x+sinx⋅cosx)]_0^{π/2} = 1/2π((π/2+0) - (0+0)) = {π^2}/4$ units$^3$

Trigonometric Substitutions

When an integral involves a quadratic radical expression, the following substitutions may be useful:

The form $√{a^2-x^2}$ may be substituted by $x=a$sin$θ$.

The form $√{x^2-a^2}$ may be substituted by $x=a$sec$θ$.

The form $√{a^2+x^2}$ may be substituted by $x=a$tan$θ$.

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