Match the parametric equations with the graphs labeled i vi...

Match the parametric equations with the graphs labeled I-VI. Give reasons for your choices. (Do not use a graphing device.) (a) $ x = t^4 - t + 1 $, $ \; y = t^2 $ (b) $ x = t^2 - 2t $, $ \; y = \sqrt{t} $ (c) $ x = \sin 2t $, $ \; y = \sin(t + \sin 2t) $ (d) $ x = \cos 5t $, $ \; y = \sin 2t $ (e) $ x = t + \sin 4t $, $ y = t^2 + \cos 3t $ (f) $ x = \dfrac{\sin 2t}{4 + t^2} $, $ \; y = \dfrac{\cos 2t}{4 + t^2} $

Transcript

00:02 So the problem says, match the parametric equations with the graphs labeled 1 to 6.

00:08 Give reasons for your choices.

00:10 Do not use a graphing device.

00:12 So we are not using graphing device.

00:15 But we know the method for one thing that we can put t equals to 0 or infinity.

00:21 Then we'll see what will be the value of x and y.

00:24 Right.

00:25 So here are few oscillatory equations.

00:30 Right.

00:32 So we will see here.

00:33 That what all oscillatory equations are there so this is oscillatory right this is oscillatory you can say i'm sorry this is fourth one right so this is oscillatory this is oscillatory and we can say this is also oscillatory right so this is how we find out right so this is also oscillatory right so now what what can we do now this is for sure that the part a and b does not contain any sinusoidal wave or sign cost kind of terms right so it will not have that value right the second third fourth and sixth values excluded from it right so now we will solve simply that we can just observe that what is happening in this problem right so so here we can see this equation, parametric equation, right? as we put t equals to infinity, right? so for x and y goes to infinity, right? if we put t equals to infinity here, so x will be infinity and y will be also infinity.

02:05 So the graph must contain one zero value anyhow.

02:12 Right? so, which is only possible in fifth graph.

02:28 Because here we can see, if you see in minutely, this is one zero value, right? this is one zero you can say here.

02:37 Here there is one zero coordinate.

02:40 Right? can we say this? so the best match is fifth graph.

02:56 Or you can say graph fine all right so now moving forward we can see here x equals to t square minus 2 t this equation is given to us this is not oscillatory we have already discussed that 1 and 5 are not oscillatory right so we can see here even we can put t equals to infinity as i have told.

03:28 So t equals to infinity.

03:31 So this is for sure after that after conclusion from above we can also directly write that the answer is one but we'll see it how.

03:43 So the equation passes through zero zero when we put t equals to infinity here right so this will be 0...

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