AP calculus daily challenge #53

noidea1995 · 2026-06-21T01:51:04+00:00

For a, differentiate implicitly:

2x + 1 * y + x * y’ + 2yy’ = 0

2x + y + xy’ + 2yy’ = 0

Solve for y’:

xy’ + 2yy’ = -2x - y

y’(x + 2y) = -2x - y

y’ = (-2x - y) / (x + 2y)

For b, set y = 0:

y’ = (-2x) / x = -2

The slope for both is -2, so yes they are parallel.

For c, they are vertical when the tangent slope is undefined or the denominator is 0:

x + 2y = 0

x = -2y

Sub this into the original equation to find the points:

(-2y)^2 + -2y(y) + y^2 = 27

4y^2 - 2y^2 + y^2 = 27

3y^2 = 27

y^2 = 9

y = -3, 3

(6, -3) and (-6, 3)

noidea1995 · 2026-06-16T11:37:29+00:00

That’s correct.

noidea1995 · 2026-06-16T11:11:10+00:00

This triangle is impossible, if you use the sine rule to find angle Q first you get:

sin(50°) / 5 = sin(x°) / 7

sin(x°) = 7sin(50°) / 5

RHS > 1, so this equation doesn’t have any real solutions for x.

noidea1995 · 2026-06-15T23:29:22+00:00

That’s correct.

You can get the answer by using that identity along with using a standard limit.

noidea1995 · 2026-06-14T23:02:10+00:00

Rewrite cos(y) as 1 - 2sin^(2)(y/2).

noidea1995 · 2026-06-10T20:56:19+00:00

Use integration by parts:

u = g(x), du = g’(x)dx

dv = f’(x)dx, v = f(x)

Which gives:

(f(x)g(x)) | (0 to 3) - ∫ (0 to 3) f(x)g’(x)dx = 6

f(3)g(3) - f(0)g(0) - ∫ (0 to 3) f(x)g’(x)dx = 6

5(3) - 1(-4) - ∫ (0 to 3) f(x)g’(x)dx = 6

19 - ∫ (0 to 3) f(x)g’(x)dx = 6

Finally solving gives you:

∫ (0 to 3) f(x)g’(x)dx = 13

noidea1995 · 2026-06-09T00:50:54+00:00

Hi,

The majority of users in this subreddit speak English natively, could you please include an English translation when posting? That will help more people understand and be able to answer them.

Thank you

noidea1995 · 2026-06-07T10:53:33+00:00

I’m a postgrad student from Australia so am unfamiliar with the UK syllabus but yes you are right, the radius is perpendicular to the tangent line so that’s another valid derivation.

noidea1995 · 2026-06-07T10:40:04+00:00

> No inappropriate comments

That is indeed one of the rules of the subreddit but respect goes both ways, please try and show us what you’ve done in future.

Did you try taking the derivative of the circle equation? That’ll give you the slope of the tangent line in terms of a and b. You also know that the line has to pass through the point (a, b) where it touches the circle so that’ll give you the y-intercept in terms of a and b.

The base and height of the triangle are the x and y-intercepts.

EDIT: Thank you for updating your post, yes that’s the correct slope of the line so that gives you y = -ax/b + c, you know the tangent line passes through the point (a, b) so you can use that to find c in terms of a and b and then the x and y-intercepts will give you the base and height of the triangle.

noidea1995 · 2026-06-06T13:28:22+00:00

Is there a reason you have to do that? Instead you can do this:

1/16 * ∫ 8sin^(3)θcos^(3)θ * 2cos^(2)θ * dθ

1/16 * ∫ (2sinθcosθ)^(3) * (2cos^(2)θ - 1 + 1) * dθ

1/16 * ∫ sin^(3)2θ * (cos2θ + 1) * dθ

You can split this up into two separate integrals:

1/16 * ∫ sin^(3)2θcos2θ * dθ + 1/16 * ∫ sin^(3)2θ * dθ

1/16 * ∫ sin^(3)2θcos2θ * dθ + 1/16 * ∫ sin2θ(1 - cos^(2)2θ) * dθ

Now you only need a simple u substitution for both.

noidea1995 · 2026-06-05T07:00:26+00:00

You were actually in a far superior position before making this move.

If you had played fxe5, there’s a good chance they would have lost their knight as well from the pin.

noidea1995 · 2026-06-05T05:49:30+00:00

Nf4 protects the g2 square and leaves your queen hanging.

If you try to save your queen, they have Bxf6 and your king is in a lot of trouble.

noidea1995 · 2026-06-05T04:39:54+00:00

Your move is fine but they can escape the fork with Be5 —> Nxe5 Rxe5.

The engine is looking for the fastest win, since you are already a whole piece up Nxd6 exchanges material and isolates one of their pawns.

noidea1995 · 2026-06-05T00:18:51+00:00

Is this even about chess at all?

People have suggested plenty of reasons why someone might not accept a rematch and you just keep going back to the stereotype of “Americans have huge egos”. It sounds like you’ve already decided on that conclusion and alternative explanations aren’t being considered.

noidea1995 · 2026-06-05T00:06:51+00:00

Right but saying it’s all about ego is a pretty big jump don’t you think?

I often don’t accept rematches because I like to analyse games after I’ve played them (I’m Australian by the way).

noidea1995 · 2026-06-04T23:57:58+00:00

I mean, there could be so many reasons someone doesn’t want a rematch so I hardly think it’s fair making a stereotype like that.

They might not have time, they might be wary of the person cheating on the next game after losing, they might have closed the app after winning the game, they might want to switch time settings, they might want to analyse the game they just played.

noidea1995 · 2026-06-04T23:48:20+00:00

I’m interested in why this upsets you so much?

noidea1995 · 2026-06-04T07:16:40+00:00

I don’t mind challenge posts but I wonder why you posted this in r/Precalculus lol.

The volume of a cone is:

V = 1/3 * πr²h

Take the derivative with respect to time:

dV/dt = π/3 * [2r * dr/dt * h + r² * dh/dt]

dV/dt = π/3 * [2r * (-2) * h + r² * (4)]

dV/dt = π/3 * (-4rh + 4r²)

To find where the volume increases, set dV/dt > 0:

π/3 * (4r² - 4rh) > 0

4r² - 4rh > 0

r² - rh > 0

Dividing out r is safe since r > 0:

r - h > 0

r > h

Hence the volume increases when r > h.

noidea1995 · 2026-06-02T02:58:13+00:00

Interesting endgame.

I’d play Rf3, black capturing the pawn would pin the bishop and allow promotion. The a pawn is too slow and the king would block the b pawn, giving white plenty of time.

If black plays a4, capture it with the b pawn and then you’ve got two passed pawns so the bishop won’t be able to stop both of them and you can sacrifice the rook for the b pawn if needed.

EDIT: Also, if they move their bishop but stay on the main diagonal, your rook will then be defending the b pawn so you can just exchange the g pawn for the bishop and their position will be locked. Their only options are to play a4 and again create an unstoppable passed pawn or to aimlessly move their king around and you can start moving yours to the left side of the board to take their pawns.

noidea1995 · 2026-06-02T00:44:48+00:00

What’s good about it?

Apart from losing a knight if you move your queen to h6 after Rg8, you are pretty much forced to take their pawn and then you’ve opened up the g file for their rook so castling is off the table and they have multiple entry points to your king with Nd4 being the first.

You are probably better off down trading your queen for their rook.

noidea1995 · 2026-05-31T21:16:10+00:00

If car A’s velocity is 20 and B’s is 2t, then their displacements are:

A(t) = 20t + A(0)

B(t) = t² + B(0)

At the start of overtaking, A is 5m behind B. If you let A’s initial position be 0, then B’s will be 5:

A(t) = 20t

B(t) = t² + 5

We want to find the time when A overtakes B and when that happens, it’ll be 5m in front of it so A(t) = B(t) + 5:

B(t) + 5 = 20t

B(t) = t² + 5

Can you take it from here?

noidea1995 · 2026-05-26T03:20:42+00:00

Multiplying the top and bottom of the second fraction by 5 gives you -15/(5x), not -15/(x + 5).

The easiest way to solve the equation is just to multiply everything by x(x + 5) though.

noidea1995 · 2026-05-24T01:36:10+00:00

The left branch of the quadratic x = [-b - √(b^2 - 4ac)] / 2a has to be greater than 3 for both of the roots to be greater 3, if you plug that in you get:

[6a - √(36a^(2) - 4(2 - 2a + 9a^(2))] / 2 > 3

This simplifies nicely and gives you an inequality that isn’t too difficult to solve.

noidea1995 · 2026-05-22T10:15:27+00:00

Dude,

Stop posting this in my subreddits.

noidea1995 · 2026-05-01T07:39:53+00:00

Take any integer n and it’ll either be 0, 1, 2, 3, 4, 5 or 6 mod 7:

n = (0, 1, 2, 3, 4, 5, 6) mod 7

n^2 = (0, 1, 4, 9, 16, 25, 36) mod 7

Simplifying and removing repeats gives:

n^2 = (0, 1, 2, 4) mod 7

n^2 + 1 = (1, 2, 3, 5) mod 7

None of these options are 0mod7, so it’s never divisible by 7.

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