Berkeley Math Tournament calculus tiebreaker

This calculus tutorial contains the solution to the 2022 Berkeley Math Tournament calculus tiebreaker! It contains a limit, derivative, and an integral at calculus 2 (i.e. AP calculus BC) level. The time limit is 15 minutes but you will also have to race with other times during the tournament. I love these math competition problems because they really force you to think and they really challenge your problem-solving skills. It's okay if you can't solve these competition problems. I, too, can't solve many of them either. It really takes lots of effort and practice in order to be good at these. Finally, I just want to tell you, to be sure to enjoy the learning process and have fun!

0:00 2022 Berkeley Math Tournament calculus tiebreaker
0:08 Q1, evaluate a limit with an integral
3:10 Q2, find the 2022nd derivative of f(x)=e^x*sin(x) at x=0
7:36 Q3, evaluate the integral of arctan(x)/x from 1/e to e
#calculus #tutorial #berkeley #blackpenredpen

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Thank you all!

[blackpenredpen]

I will be there on Nov 4th, 2023 on UC Berkeley campus!

[sethb]

For the second problem, if you differentiate e^x*sinx 4 times, you end up getting -4e^x*sinx, meaning that every 4 derivatives, you're just multiplying by -4. Notice that 2022/4 is 505 with a remainder of 2, meaning we're going to be multiplying by -4 505 times, then differentiating twice more. So that gives us (-4)^505*e^x*sinx. If we differentiate that twice more, we get (-4)^505*2*e^x*cosx, which is our 2022nd derivative. Now we can plug in 0 for x, which leaves (-4)^505*2. We can simplify this a bit to get (-2^2)^505*2=(-1)^505*2^1011. Recall that -1 to an odd power is simply -1, so the answer is just -2^1011

[brianxx]

Half my life ago I had the kind of mind that would have spontaneously started solving these. Now I sit and enjoy someone else's solution. I feel old but I am glad you exist!

[monishjain]

At 10:50, we may use kings property and formula for arc tan a + arc tan b that would give arc tan infinity put (pi/2) there and just solve. Quite a bit easier that way

[Ricardo_S]

On Q2 I found the derivative

-2¹⁰¹¹cos(x)e^x

Evaluating at 0 effectively gives -2¹⁰¹¹

[nathandaniel]

I often overlook these videos, just assuming that these problems would be faar too difficult, but I'm pleasantly surprised that I can follow along easier and figure out the answers myself.

[RyanDalzell-lmjo]

This is really cool man, I’m currently attending UC Berkeley right now and I think it’s awesome that you sponsored our Math Tournament as well as graduated from here!

[LmaoDed-haha]

In the third question, another method would be substituting
X= 1/t
dx=-1/t² and proceed you will get answer π/2 easily.

[hassanalihusseini]

The last one I never would be able to solve. Thanks for the video!

[ascanius]

I had another solution for the 2022 problem I set S=e^x sinx and noted that d^4/dx^4S= (-4)S. Then you can do the 2020 differentiations and get a factor of (-4)^2020 and do the last two differentiations by hand. I like your way more. At times the complex way ist just the simpler way;-)

[davidcroft]

The third question is actually simpler: you divide your integral in due part ( I=1/2*(I+I) ) and in the second you substitute t=1/x. After some calculations, the 2nd integral is similar to the 1st one, except arctan(x)-->arctan(1/x); if you put them together you have Now we know that sum is identically pi/2 (if x>0), therefore I=pi/4*∫1/x*dx, and after elementary calculations you obtain I=pi/2

[krishgarg]

i was able to do 1st and 3rd but the using complex number for 2nd was beautiful.

[rayquazabtc]

For the first question we can use Lhospital directly it is infinity/infinity indeterminant form to differentiate than integral we can use Newton leibniz theorem

[manasijbhattacharjee]

For the third question, we can also proceed with another method from third step... we can write it as (it is a property of DI) ∫ 0 -> 1 (f(x) + f(-x)) dx then you will get something like arctan(x)+ arccot(x) which is equal to ∏/2.. ( I couldn't find the right pi 😅.. and yeah x here means the e to the power u.... Btw, noice integrals.. Would love to see more of such covered in your future videos.. 😎

[bruhnish]

me: struggling with second derivative
meanwhile this guy: doing 2022nd derivative in 2022 milliseconds

[jordonludlam]

My guy really said I don’t need all 15 minutes. I can do it and add in a sponsor for a math comp

[abhinavs]

Please Make a Playlist with Competition Math problems ❤

[henrybarber]

For the third integral, you can make the substitution u = 1/x. You find that the integral is equal to that of arccot(x)/ x on the same interval. So the integral is equal to 1/2 of ∫(arctanx + arccotx) / x dx. Then you just need that arctanx + arccotx = π/2 to find the answer quite simply.

[Dreamprism]

Great video! Good explanations. I'm glad for you and everyone attending the BMT.

[nanamacapagal]

For the second problem, that one Michael Penn video about linear algebra with derivatives was still fresh in my mind, so I'll be using that.

Notice that when you differentiate a linear combination of e^x*sin(x) and e^x*cos(x) you get another linear combination of the two functions.
Let the first number in the vector space be the e^x*sin(x) terms and the second the e^x*cos(x) terms.
Differentiation can be represented by this matrix:
D = [[1, -1]; [1, 1]].
The starting vector is S = [1; 0].
That means we need to find D^2022 * S.

Usually when a big power of a matrix shows up it's a good idea to diagonalize:

Eigenvalues: (1 + i), (1 - i)
Eigenvectors: [1; i], [1; -i]
D = A'XA
= [[-i, -1]; [-i, 1]][[1 + i, 0]; [0, 1 - i]][[1, 1]; [i, -i]]
D^2022 = A' * X^2022 * A
= A' * [[(1 + i)^2022, 0]; [0, (1 - i)^2022]] * A
= A' * [[-i(2^1011), 0]; [0, i(2^1011)]] * A
I'm in a bit of a rush but the next step should be multiplying out A' * X * A * S to get the final result


Note: I write my matrices inline like this:
A = [[R1C1, R1C2, ...]; [R2C1, R2C2, ...]; ...]. Commas separate elements in rows and semicolons separate rows.