Parametric equations hw help pls

[brazen jay]

Hello could anyone help me out been given this hw but havent been taught it in detail yet would very much appreciate links to videos or just the method

[halcyon nebula]

I haven’t been taught it either but I gave it a go

Tried rearranging at first but then just found derivatives

Q1 that is

[worn nest]

Not sure what you were doing at first, but then you calculated dx/dt, dy/dt and dy/dx correctly, as well as the equation of the tangent.

[halcyon nebula]

Q2

[halcyon nebula]

Yeah I tried rearranging thinking I could find the equation of the curve

[worn nest]

Well, that might be possible sometimes, but certainly not always.

Besides, I think parametric derivatives are simpler than implicit ones.

[halcyon nebula]

Okay dokey

I’m just doing these by intuition. Don’t know why school leave this as one of the last topics

Green writing Q3, not so sure about this one.

[worn nest]

Yeah, seems good.

[halcyon nebula]

Q4 in green. Left as spoiler so you can have a go yourself first after looking at first few Q’s @brazen jay

Not much room left. Q5, same as above

[worn nest]

Alright, let me try first, then.

Sorry, need to leave for a bit, will do this in 15-20 minutes.

[worn nest]

Alright, I'm back.

4a.
i.
sin(2θ) = 2sin(θ)cos(θ)
cos(2θ) = 2cos(θ)^2 - 1 = cos(θ)^2 - sin(θ)^2 = 1 - 2sin(θ)^2
ii.
cos(θ) = 3/5
θ = ±arccos(3/5) + 2πn
For 0 < θ < π/2 the only root is θ = arccos(3/5).
sin(2θ) = 2sin(arccos(3/5))cos(arccos(3/5)) = 2√(1 - (3/5)^2)(3/5) = 2(4/5)(3/5) = 24/25
cos(2θ) = 2cos(arccos(3/5))^2 - 1 = 2(3/5)^2 - 1 = -7/25

4b.
i.
x = 3sin(2θ), y = 4cos(2θ)
dx/dθ = 6cos(2θ), dy/dθ = -8sin(2θ)
dy/dx = -(4/3)tan(2θ) = -(16/9)x/y
ii.
Using the values from 4a for our point θ = arccos(3/5):
(x, y) = (3sin(2θ), 4cos(2θ)) = (72/25, -28/25)
dy/dx = -(16/9)x/y = (16/9)(72/28) = 32/7
The equation of tangent is:
y = (32/7)(x - 72/25) - 28/25

x = 2t + 4, y = t + 6
Let's find y(x).
t = x/2 - 2
y = x/2 - 2 + 6 = x/2 + 4
So:
∫(ydx, 0, 6) = ∫((x/2 + 4)dx, 0, 6) = (x^2/4 + 4x)|(0, 6) = 36/4 + 24 = 33

[worn nest]

Did you get the same results?

[halcyon nebula]

Yep 🙂

[halcyon nebula]

I just wrote cos2theta as cos^2theta - sin^2theta but I can ofc sub in further to get just cos^2 and sin^2

[worn nest]

I think you got something different in 4bii.

[halcyon nebula]

I just subbed in my theta value into 4bi to find dy/dx

[worn nest]

Ah, I see your mistake. You accidentally wrote tan(θ) instead of tan(2θ).

[brazen jay]

I got that for 4

[worn nest]

Well, I don't like rounding, but I got the same result, essentially. Though, you did make one mistake: at the end, you should have -100/7, not +100/7.

[halcyon nebula]

Thank you haha

[worn nest]

You're welcome!

[halcyon nebula]

I’m missing a second coordinate and it’s because I need a plus minus inside the brackets. Why is that?

[worn nest]

(8cosh(x) + 1)(cosh(x) - 4) = 0
1.
8cosh(x) + 1 = 0
cosh(x) = -1/8
No solutions.
2.
cosh(x) = 4
x = ±arcosh(4) = ±ln(4 + √(15))
So, what you forgot is that the equation cosh(x) = a (a ≥ 1) has solutions x = ±arcosh(a), not just arcosh(a).

After all, cosh(x) is even.