A: 3
B: π
C: 3π
D: π/2
First, lets look at some graphs of our x and y values, cosine and sine.
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To find our starting point on our parametric graph of x and y (when t = 0), we can use these two graphs to quickly tell that when t = 0, x = cos(0) = 1 and y = sin(0) = 0. Now we start drawing our graph:
Great start! Now where do we go from here? We have to determine the shape that we are drawing. Since we're dealing with sine and cosine, let's try to remember from our Trigonometry class:
If we imagine that θ gets smaller and bigger between 0 and π, notice that the length of sin(θ) follows along the edge of the circle. The same goes for cos(θ). This means that the shape of our parametric graph will follow a circle.
Let's continue drawing our graph to the next obvious point. By following the line on our cosine graph to halfway between 0 (our start point) and π (our end point):
we see that x = cos(π/2) = 0.
Following our sine graph:
we see that y = sin(π/2) = 1.
We can now say that the half-way point of our parametric graph is (x, y) = (0, 1). Graphing this out, we get:
Which confirms that the curve follows a circle.
Following the curves of sine and cosine from π/2 to π will give us:
and
We now have the end point of our parametric graph of (x, y) = (-1, 0).
Now, in order to find the length of this curve (our original goal), let's notice that what we have so far looks like the top half of a circle, which it is (since we are following the curve of a circle)! If we fill out the missing part of the circle:
we can now use basic geometry to find the length of our parametric curve!
We know that a circle has a circumference of 2πr where r is the radius of the circle. Since the radius of our circle is 1, our full unit circle has a circumference of 2π. We, however, only want the top half of that unit circle, giving us 2π/2 = π.
B, therefore, is the correct answer.
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