Me & my friend have spent 2 hours on these 2 questions. Area with parametric equations. Reddit Calc2

We will find the area under the curve when given parametric equations. This question is from Reddit r/maths and will help you with your Calculus 2 class. / 69obx32zfj
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#calculus #bprpcalculus #apcalculus #tutorial #math

[@bprpcalculusbasics]

The second question: ruclips.net/video/xbkqsDCigt4/видео.html

[@cyrusyeung8096]

The video is private, and is not viewable.

[@xinpingdonohoe3978]

y dx = y dx/dt dt.
That's the good thing about Leibniz notation. Everything that you think should work, does work, and it helps to keep some stuff intuition based instead of just memorisation all the time.

[@RadChromeDude]

yessssssss this makes parametric equation so easy!

[@TheEricthefruitbat]

I love the red herring by the test givers: the point A has nothing to do with the problem/solution.

[@roland3et]

Nice solution. Nevertheless, I found the other one (mentioned at the very beginning and unfortunately wiped away immediately) much more simple, and therefore easier to compute: No reason for changing dx to dt, nor any need for recalculate the limits of the integral!
And, furthermore, Steve's argument, this first easy way wouldn't work in general, is canceled out by himself a minute or two later, using pretty much the same conversion to find the new limits of the integral in t.
Despite of that little dispute, it's again a great video of pbrp, with brilliant explanations!
Love all your channels, Steve 👍!
🙂👻

[@orangeflow9809]

I love this channel so much

[@uyangapuujee4508]

Almost 1 month left until my first a level maths exam. I’m still learning some new tricks from you 😭😇

[@manyifung5411]

May you do the integration of ln(x)W(x)?

[@aleksandarcvetkovski1784]

Best teacher ever ❤

[@borismedved835]

Should also be a first grade English lesson here. "My friend and I have spent..."

[@TVETHigh]

Thumbs 👍 up, best explanation my good Sir.

[@btb2954]

inverse of (1-x/2)=inverse of (2^y-1) get the inverses, solve for y then integrate

[@norfolknchance657]

"My friend and I"

[@Mecha_Math]

Very good 🎉🎉❤

[@Chrisuan]

Question: how do you get the bounds of integration in t, if the x-equation isn't solvable for t?

[@carultch]

Here's my own example, following Paul's method.
Given:
x(t) = cos(t) - t
y(t) = t*e^(-t)
Interested in the area in the first quadrant of the parametric curve
The limits of t are:
t = 0 to 0.74 (no exact value exists for the root of cos(t) - t)
Area under curve:
integral y(t) * x'(t) dt
Carry out derivative for x'(t):
x'(t) = -sin(t) - 1
Multiply with y-function to get integral:
integral -(sin(t) + 1)*t*e^(-t) dt = 1/2*e^(-t)*[2*t + t*sin(t) + (t + 1)*cos(t) + 2] + C
Evaluate at t = zero:
1/2*e^(-0)*[2*0 + 0*sin(0) + (0 + 1)*cos(0) + 2] = 3/2
Evaluate at t = 0.74:
1/2*e^(-0.74)*[2*0.74 + 0.74*sin(0.74) + (0.74 + 1)*cos(0.74) + 2] = 1.25574
Subtract to get our final answer for area:
0.244

[@Robin-Dabank696]

​@@carultchso short answer: use a calculator

[@carultch]

@@Robin-Dabank696 True. It is solvable in theory with the methods before calculators existed, but no exact solution exists to that example. I realize my comment with a link disappeared, so "Paul" is a mystery. Paul is Paul Dawkins of the website Paul's Online Notes.

[@javier_me2]

Second comment :D

[@robertpearce8394]

"My friend and I"

[@andyworthen6587]

This is math, not English.

[@justarandomnerd3360]

Actual 🤓

[@rafastyles1002]

Me and the boys

[@Thenerdycat61]

Bro this is calculus not English 💀💀💀