Jill Cockerham

Vector Functions

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Example 1

a right triangle representing initial velocity

Suppose a frog is fired from a cannon at ground level, at an angle of 60 degrees to the ground, with an intial velocity of `100` ft/s. Suppose we would like to represent the position of the frog by a vector function, `vec p= < x, y >`, where `x` represents the horizontal distance the frog has traveled and `y` represents the height of the frog. Both `x` and `y` are functions of time, independent of each other.

We can break down the inital velocity into its initial horizontal and its initial vertical components, using trigonometry. By special right triangles, we see that the initial horizontal velocity is `50` ft/s and the initial vertical velocity is `50 sqrt(3)` ft/s.

Galileo discovered that horizontal velocity of a projectile remains constant (ignoring air resistance, at least). Thus, the horizontal speed of the frog remains at a constant `50` ft/s for his entire flight. Thus, the horizontal distance the frog has traveled at time, `t` is given by `x=50t`.

animation of the frog's position

The horizontal velocity of the frog, no doubt, changes during his flight. This is because of acceleration due to gravity. From calculus, one can derive the frog's height, `h(t)` at time `t` to be `h(t)=50 sqrt(3)t-32t^2`.

We now know the frog's horizontal position in terms of time and his height in terms of time. Therefore, `vec p= < 50t, 50 sqrt(3)t-32t^2 >` is the vector function that outputs a vector which points from the origin to the frog's position at time `t`.

Example 2

animation of a unit circle drawn by the vector function

There are some functions that are difficult to represent in `y=f(x)` form, such as the unit circle. It must be split into two separate functions to even be represented in `y=f(x)` form, since it is not a function. This is a good example of a funciton that is much easier to represent in parametric form. In this example, of course, we will look at it in vector function form.

If we let `vec v= < x(t), y(t) >` where `x(t)=cos(t)` and `y(t)=sin(t)`, we get a vector function, whose vectors point to points on the unit circle. By plugging in various values of `t`, you can see that as `t` increases, the vector moves counter-clockwise around the circle. In the example to the right, `t` is between `0` and `2 pi`. But since since `cos(t)` and `sin(t)` are cyclical, the vector function repeatedly draws out the unit circle for all values of `t`.

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