Question

Can someone please verify my answer to this Calc II parametric equation question?

Answer 1

I think part b is correct, because when sin is 0 or pi, the result is zero. Thus, the denominator is zero, making the tangent line vertical.

Answer 2

Yes for 4b
sin(t^2) = 0

=> t = 0 or sqrt(pi)

Answer 3

Note that \(\large\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\). You plugged them in the reverse order! :-/

Answer 4

See... this is why I like peer review. Haha.

Answer 5

So the answer for 4a would now be this, correct?

Answer 6

And the new answer to 4b would be:

$$\frac{\sqrt{\pi}}{\sqrt{3}}$$

since

$$\frac{3}{2} \cdot \frac{\sqrt{\pi}}{\sqrt{3}} = \frac{\pi}{2}$$

and
$$cos(\frac{\pi}{2}) = 0$$,

so the tangent line is vertical.

Answer 7

And since $$\frac{3\pi}{2} > 2$$
I don't need to consider it.

Answer 8

But \(\sqrt{\pi}\approx 1.77245 <2\), so you need to consider that too!

Answer 9

for for 4a: besides the reversed order, I am also getting a minus sign in the derivative, since cos' = - sin

Answer 10

@b87lar but you don't need to do any differentiation for part (a) since they were so kind as to provide you the derivatives.

Answer 11

$$dy/dt$$ and $$dx/dt$$ is already given

Answer 12

oh you'right - sorry!

Answer 13

:) np

Answer 14

@LolGoCryAboutIt your answer for (a) is correct now, but for (b), it's \(\large t=\sqrt{\dfrac{\pi}{3}}\) and \(\large t=\sqrt{\pi}\).

Answer 15

Oh! and @ChristopherToni, I see what you mean.

t can also equal to $$\sqrt{\pi}$$
because
$$(\sqrt{\pi})^{2} \cdot \frac{3}{2} = \frac{3\pi}{2}$$

Answer 16

Exactly; and furthermore \(\large \sqrt{\pi}\leq 2\). :-)