Roark

It's from teach yourself calculus. The chapther is an introduction to diff eqs and the section is separation of variables. Here's a reply from Karl of Karl's calculus forum. It sounds like you are both on the same track but that track is down the line for me. I'll be getting to this in about six months. Thanks. Reply to Peter: It would take a lot more than 3 weeks to do this diff. eq. by separation of variables because this one can't be done that way. This one you have to first find the integrating factor that makes it exact. Then you apply the method for solving exact equations. There is no recipe for finding integrating factors. You have to do it by educated guessing. After about two or three guesses I found that the integrating factor for this one is 1/y^2. The equation becomes (1 - x^2/y^2)dy + (2x/y)dx = 0 Notice if you take the partial derivative of the left-hand factor with respect to x, it is equal to the partial derivative of the right-hand factor with respect to y. That is what it means for an equation to be exact. This means that I simply integrate either factor -- either the left dy or the right dx to get one step closer to the solution (whose form will be F(x,y) = C). I'll do the right. F(x,y) = x^2/y + g(y) Notice that instead of C as the constant of integration, I used g(y). Why? Because y is independent of x, and I integrated dx. So a function of y is a constant with respect to x. Now if I take the partial derivative of this solution with respect to y, it should be equal to the left-hand factor ¶F = -x^2/y^2 + g'(y) = 1 - x^2/y^2 ¶y You get some cancellation and find that g'(y) = 1 or equivalently g(y) = y. So the solution is F(x,y) = x^2/y + y = C Now multiply by y and you have x^2 + y^2 = Cy Hope this helps.

This doesn't match the answer in the book. Did we figure that the book is wrong?

Got it! The rate of change of the area of the rectangle is -7 only when L=6. Incidently, the rate of change of the W is 1.5 only when L=6. How would you go about testing it?

Interesting. I assumed that the circle would have to be shrinking as the length of the rectangle decreased. You're assuming that the circle size is staying fixed and, after re-reading the problem, I think you're right. The radius of the circle should stay at 5. Indeed, in you're solution you fixed the radius at 5 instead of leaving it variable. But there's a problem. The answer doesn't check out. Consider: t0:l=6, w=8, A(rect)=48, r=5, d=10, dl/dt=-2 t1(one second later):l=4, A(rect)=48-7=41, w=41/4=10.25, d=11, r=5.5 For the area of the rectangle to be decreasing at -7, the circle would have to be getting bigger. What is the correct answer if the circle doesn't change?

Hi, you mixed up w and l but I get it. Express w in terms of l and differentiate since you know dl/dt. Here's what I don't get: A rectangle inscribed in a circle looks like this: ([]). We know that when l=6, w=8 and A=48. If l=6 and is decreasing at a rate of -2/sec then after 1.5 seconds: l=6-(1.5*2) = 3. If A is decreasing at -7/sec then after 1.5 secs A=48-(1.5*7) = 37.5. This means w = 37.5/3 = 12.5 which means the area of the circle is getting bigger. What in this problem tells you what is going on? Just because the length is decreasing at a certain rate, the width could be increasing or decreasing at any rate. I assumed that the rectangle was shrinking in porportion to itself. I drew both diagonals and kept the inner angles constant. It looks like you kept the diagonal constant yet after 1.5 secs it has grown to 12.855. What's up with that? ...and thank you.

Hi, I've spent a week on this problem, and I think the answer in the book is wrong. Here it is: A rectangle is inscribed in a circle of radius 5 inches. If the length of the rectangle is decreasing at the rate of 2 inches per second, how fast is the area changing at the instant when the length is 6 inches. HINT: A diagonal of the rectangle is the diameter of the circle. The answer given is -7 sq in/sec. It is not clear whether you are to find the rate of change of the area of the circle or the rectangle. First consider the rectangle: The width is 8 making the area 48. After 1.5 seconds the length is 3, width is 4 and area is 12. So the answer given of -7 in/sec can't be right since 48 - 21/2 does not equal 12. Consider the circle. Initially the area is 25pi. After 1.5 seconds the radius is 5/2 so the area is 25/4pi. Again the answer given can't be. For the rectangle I get an answer of -32 sq. in/sec. This bears out with my post 1.5 second test. For the circle I found dr/dt to be -5/3 and dA/dt = -50/3. The dr/dt bears out as well, but the dA/dt can't be right since after 1.5 seconds -50/3 * 1.5 = -25 and 25pi - 25 does not equal 25/4pi. I'm stumped. Thanks in advance for any help.

Hi, I've finished "Teach Yourself Calculus" (in only 11 months!) and I feel like I've got a good foundation (Equivalent to Calculus 2, I think) for what's next. So...what's next? I'd like to get to Stochastic Calculus for Finance and Game Theory. I'm tempted to get a book on differential equations and/or partial derivatives. I think I'm lacking a big picture outlook of what's out there, how it's related, and where to go next. Thank you in advance for any advice.

Don't sweat wanting to just divide 697 by 7 to get the answer. I abbreviated the question from the book. Had I put it in verbatim, you wouldn't have gone there. -Peter

Cool. I don't understand anything after f(x)= ... but cool! I'll have to wait till my self-training catches up with your answer but thank you.

Your missing something. 679 / 7 = 97 candles that you can then burn to make 97 stubs, those 97 stubs make 13 more candles with 6 stubs left over...and so on.

I Love Math! My daughter, age 9, has a book of Mensa puzzles for kids. She hit me with this one: A guy recycles candle stubs and can make 1 new candle out of 7 stubs. How many candles can he make out of 679 stubs? The solution is simple: Stubs, Candles Made, Leftover Stubs 679, 97, 0 97, 13, 6 19, 2, 5 7, 1, 0 Answer: 113 So, I’m teaching myself calculus and I know that I can find a general solution for this problem. Well…not quite. I know that the answer is a series: f(x) = { (x/(7^1)) + (x/(7^2)) + (x/(7^3)) + ... + (x/(7^n)) } Testing the series with some actual values I get: x, f(x) 679, 113.120 15,139, 2,552.103 3,020,302, 503,174.011 (The fractional portion is the stub left over after burning the last candle) What’s the general solution?

That's the equation that's given in the book. Is this a valid approach? (y^2 - x^2)dy - 2xydx = 0 (y^2 - x^2)dy = 2xydx (y^2 - x^2)dy/dx = 2xy (y^2 - x^2)dy/dx = x -------------- 2y Intergrate both sides: (y^2 - x^2)/2y= 1/2x^2 + C Intergrate both sides: (y^2 - x^2)= yx^2 + C (somehow reduces to?) Ans: x^2 +y^2 - cy = 0

Dude, If you're bored then please take a look at my differential equation problem. I'm having trouble finding a good resourse on the net that lays out an introduction to differential equations and is full of worked problems. If you know of one, please share. Thank You!

I think I got it but I want to dbl-check. When do you need it?

Hi, I'm just starting to do differential equations and I'm stuck on this one: (y^2 - x^2)dy - 2xydx = 0 Ans: x^2 +y^2 - cy = 0 How do I separate the variables? What's the intergrating factor? Thanks!

Thanks! I actually got it last night. It seems like cheating when you put c = ln c. I know c can be anything but isn't the answer now only true if c = ln c?

Hi, Here's the problem: dy/dx = y/a The answer is y=ce^(x/a) I can get to : x = a ln |y| + c and I know that x=ln y is the same as y = e^x but I can't make the connection to the answer. Thanks.

You don't "touch Mr. Calculus." You command him to vanquish you enemies and hope like hell that he doesn't turn on you. Best thing to know: "The definite integral is a sum of an infinite number of algebraic products, one with a factor of which, in the limit, becomes infinitely small." This is for things you can graph, likes area and arc lenghts. But in the abstract, the known and potential applications of calculus are immense.

Thanks. I've restated the problem in a different thread. Please check it out.

Here's the problem: Find the length of an arc of a the curve y = ln x between the points where x=1 and x=2. (To carry out the integration rationalise the numerator.) So: y=ln x, y'=1/x Length = I(sqrt(1+( 1/x^2) )dx (I for integral...how do you get math symbols?) "...rationalise the numerator...": sqrt(1+( 1/x^2)) = x^2 + 1 -------------------- x sqrt(x^2 + 1 ) or x^2 + 1 --------------------- sqrt(x^4 + x^2 ) Answer: sqrt(5)-sqrt(2) + ln ( ( sqrt(5) - 1 ) / 2(sqtr(2) - 1) ) How do you integrate it? Is there a standard form? From the answer it looks like they used: I(sqrt(u^2 + a^2)du = 1/2 u sqrt(u^2 + a^2) + 1/2 a^2 ln (sqrt(u^2 + a^2)) Thanks!

Hi, How do I integrate Sqrt( 1+(1/x) )? Thanks.