- g(t)=(cost2,sint2),t∈[0,π]
Notice that ∣g∣=1 for all t and π2>2π
Then the image of [0,π] under g is indeed the whole unit circle
- g(t)=(cosht,sinht),t∈[−2,2]
We'll notice that cosh2t−sinh2t=1 for all t
So the needed sketch is just a part of the hyperbola x2−y2=1 with endpoints at (cosh(−2),sinh(−2))and(cosh2,sinh2)
- g(t)=(2−3t,5t+4),t∈[0,1]
Both coordinates depend linearly on t so the needed sketch will be a line segment with endpoints at g(0) and g(1)
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