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

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bprp calculus basics

Күн бұрын

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@bprpcalculusbasics 9 ай бұрын

The second question: kzbin.info/www/bejne/rpPOoqZ6eM6aqpY

@cyrusyeung8096 9 ай бұрын

The video is private, and is not viewable.

@xinpingdonohoe3978 9 ай бұрын

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 8 ай бұрын

yessssssss this makes parametric equation so easy!

@roland3et 8 ай бұрын

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 9 ай бұрын

I love this channel so much

@TheEricthefruitbat 8 ай бұрын

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

@btb2954 8 ай бұрын

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

@randomxxvids 8 ай бұрын

Best teacher ever ❤

@uyangapuujee4508 9 ай бұрын

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

@manyifung5411 9 ай бұрын

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

@borismedved835 8 ай бұрын

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

@TVETHigh 9 ай бұрын

Thumbs 👍 up, best explanation my good Sir.

@Chrisuan 9 ай бұрын

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

@carultch 9 ай бұрын

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

@naming_is_harddd 8 ай бұрын

@@carultchso short answer: use a calculator

@carultch 8 ай бұрын

@@naming_is_harddd 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.