Question

Calculus help for a medal?
(posting problem below)

Answer 1

I really don't know where to start.

Answer 2

The slope field tells you the slope of a tangent to the curve at any of those points, with the little hash mark line. You can not cross any of those, and the curve will follow the field lines.. i think the first part you just have to do a general sketch of the thing.
Notice the slopes at x=2 and x=0 and x=+2 are horizontal or zero.

Answer 3

looks like you intercept the y axis beteween y=1 and 2,

Answer 4

To actually solve for the function from that DE , it is straight forward, you can separate variables y on one side and x on the other

\[1 *dy = x \sqrt{4-x^2}*dx\]

just integrate both sides of that now, and you get y=f(x)

Answer 5

the integration constant can be had from the given point (2,2)

Answer 6

Hey sorry I just got back to my computer.

Answer 7

is this th ebeginning of diff eq course, or a calc question?

Answer 8

I'll read it over now.

Answer 9

I'm not sure I understand your question.

Answer 10

"is this th ebeginning of diff eq course, or a calc question?"

What do you mean by "diff eq"? I am in AP calculus if that helps.

Answer 11

differential equations, or just a prob from calc book class

Answer 12

yeah so each little line in the field tells you the value of the derivative, slope of the tangent to the curve at that point..

you can draw any family of curves onto that graph, they want the one through (2,2)

Answer 13

It is a practice problem from my calculus text book.

Answer 14

Sort of like those iron filings in a magnetic field, they tell you the direction of the field at any little section

Answer 15

Okay so I would need to identify one of the lines that is shown in the field that come into contact with (2,2)?

Answer 16

yes, following the slope field, dont cross any of those little lines and follow the slopes

Answer 17

so that is part a. What should I do for part b?

Answer 18

following the field, looks like you should intersect the y axis between 1 and 2 to stay in the field lines, then mirror back up to (-2,2) on the other side

Answer 19

This is what I got for a

Answer 20

since this is calc and not DE, it wont use any special techniques to solve the thing, just integration you are used to. Remember if you integrate a derivative, you arrive at a family of functions, (that is what the +C constant is ) you can have many functions.

Here just separate the variables and integrate both sides.

\[\int\limits 1 *dy = \int\limits x \sqrt{4-x^2}*dx\]

Answer 21

looks fine for the sketch, you can start anywhere on the graph and make a similar sketch, (that is what the constant from integrating comes in)

Answer 22

the initial condition there, was start at (2,2), you can start anywhere

Answer 23

So for integration I got \[\frac{ -(4-x^2)^\frac{3}{2} }{ 3 }\]

Answer 24

+ C

Answer 25

if you integrate that you arrive to

\[\huge y = \frac{ -(4-x^2)^{3/2} }{ 3 }+C\]

yep

Answer 26

So now we plug in the (2,2)?

Answer 27

to get the C value, you can use the initial point (2,2) put that in for X and Y, and solve for C, you get C=2

so the "particular solution" is
\[\huge y = \frac{ -(4-x^2)^{3/2} }{ 3 }+2\]


ha, yeah you are aahead of my typing, good

Answer 28

to get a feel for the slope field,
https://www.desmos.com/calculator

graph that y(x) with the constant C in it, and add the slider, see how changing C moves the graph up and down..

Answer 29

the idea, that that differential equation dy/dx = ...
has a 'family' of solutions, not just one

Answer 30

the graph is a field like that

Answer 31

So the graph is the field representing the "family" of solutions, and by plugging in the given values, we find what C equals to which gives us the specific equation we are looking for?

Answer 32

yea, that is why they always want you to add + C on integrations

Answer 33

think of it as a magnetic field or something, what the path of a magnet would do depends on where you drop it initially

Answer 34

here it is the same general shape just a shift in Y for the path, but it can be any kind of field

Answer 35

And that is all the problem asks for?

Answer 36

yeah, long as you get the concepts , i guess, i think these things is where math actually gets fun , not just a machine solving integrals..lol

Answer 37

if you really like math..
MITOCW , i think the guy is Arthur matuck or something differential equations, good vids to learn this type stuff, i used it retricea second source while taking the course a few years back and it helped out a bunch.

Answer 38

goes into much more being MIT than my course did, but still good stuff

Answer 39

Honestly, I've always loved math, but calc has been a pain. Thank you for the sources. I'll look into them. :)

Answer 40

yeah calc can suck at times, once you get decent at that , here is the fun stuff
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/

Answer 41

have fun, goodluck

Answer 42

Thank you!