Question

A fun problem, I don't know if I've asked it before here or not.

Find the function such that tangent of this function is the slope of a tangent line to this function.

Answer 1

Deja vu!

Answer 2

Haha well I guess that's a yes. =P

Answer 3

I'll drop a hint : homonyms

Answer 4

I'm not sure what you mean by "tangent of this function"?

Answer 5

Tangent of a function is the same as sine of the function divided by cosine of the function =P

Answer 6

Hmmm interesting indeed. What kind of class is this for?

Answer 7

This isn't a class, just a fun question I came up one time when I was bored. It's a fairly simple differential equation that's sort of a play on words.

Answer 8

Ahhh that's why I don't get it.... Haven't taken differential equations yet.

Answer 9

It's basically just a calculus question though, nothing too complicated. \[\Large \dfrac{dy}{dx}=\tan(y)\] Solve for y. So this is a separable equation. \[\Large \int\limits \ \frac{1}{tan(y)} dy = \int\limits dx\] since tangent is just sine over cosine, we have it's own derivative there. \[\Large \int\limits \frac{\cos (y)}{\sin (y)}dy=\int\limits dx\] Just solve these integrals and we get: \[\Large \ln| \sin(y)|=x+C\] Now we just solve for y \[\Large y=\sin^{-1}(Ke^x)\] Here I just replaced C with another arbitrary constant K=e^C.

So that's all there is going on there, basic calculus 2 sort of stuff.

Answer 10

A common question to solve on your own is, what function is its own derivative? \[\Large y= \frac{dy}{dx}\] Now just multiply dx onto the other side and divide y to the other side to get two integrals. And then you can solve away for y. You should get something fairly predictable. =)

I can explain anything you'd like to learn more of, but from what I gather you've taken at least calculus 1 correct?

Answer 11

haha that's exactly correct. I'm going to be taking calc 2 next semester. I most definitely appreciate your offer of help. I will keep you in mind, friend! I can always use more help.

Answer 12

Yeah definitely! I like introducing people to this, so just out of curiosity, do you know what the derivative of this is? \[\Large A(r)=\pi r^2\]

Answer 13

\[dA/dr=2\pi r\]Correct?

Answer 14

Yeah, why do you think the derivative of area is circumference?

Answer 15

Because as the radius grows or shrinks, the circumference grows or shrinks in relation to the radius? Not exactly a rigorous answer lol

Answer 16

Well, it's kind of like an infinitesimally thin layer of area around the entire circle at that radius is what a circumference is.

Answer 17

Very interesting. So as the circumference increases by dr, we add on an infinitely small ring around the circle. And this ring is nothing but a rectangle with a width of dr, and length 2 pi r. I think I understand, but not quite a solid grasp yet.

Answer 18

Reminds me of solving for volume of solids of revolutions.

Answer 19

Yeah, exactly the same idea. So then it follows that since it has a width of dr then we can add up an infinite number of these infinitely thin slices to get Area.

So you might want to try deriving the volume of a sphere, and from that derive the formula for surface area of a sphere.

If that's not hard enough or you're having fun try finding the surface area and volume of a doughnut shape.

Answer 20

Oh yes I have done the volume of the sphere and the doughnut shape. I found it very satisfying that the volume of the sphere came out to 4/3 pi r^3. I definitely remember now how we add on layers, as in slices or discs/washers, to the solid to find the volume. We ended the semester doing problems involving work like pumping water and with Hooke's law.

I'm actually excited to start calc 2 and learn about inverse function stuff and derivatives of logarithms.

Answer 21

Ahh awesome. Well I'm sure you'll be fine.

Answer 22

Thanks! You'll be hearing from me when my brain starts hurting.

Answer 23

I really like power series, integration by parts and definitely trigonometric substitution. A lot of fun stuff coming up! =)

Answer 24

I hear from former students of calculus that they HATED Fourier series and the like. What do you make of that?