mm-1979 I have one last question: in the definition of sinh(x) and cosh(x) is the x the angle between the P=(cosh(x),sinh(x)) ? If is it so, who do I prove it? >I have one last question: >in the definition of sinh(x) and cosh(x) is the x the angle between the P=(cosh(x),sinh(x)) ? If is it so, who do I prove it? Let x = cosh(t) and y = sinh(t). So the parameter t determines x and y. The angle theta to which you refer is the standard polar coordinate angle, and we know tan(theta) = y / x = sinh(t) / cosh(t) = tanh(t) so theta = arctan( tanh(t) ). In particular, no, theta does not equal t. --Lynn > in the definition of sinh(x) and cosh(x) is the x the angle between > the x-axis and the line which intersects the hyperbola (in the point > P=(cosh(x),sinh(x)) ? No, it's not. -- Rich Carreiro rlcarr@animato.arlington.ma.us Is there any resource on the webb where I can find formulas to the most common fractals, like the Julia? I only need the raw formulas, like for instance the Mandelbrot formula z= z*z+c When I have that, I can program the rest. My company is developing animated wallpaper for mobile phones. These are gif-animations and I found that fractal zoomings are very suitable for these. So far I have only used the Mandelbrot, but I would like to produce much more wallpaper based on different formulas. > presented with this Question at Uni and I could do with some help on how > to go about solving it: A company that manufactures computer housings is trying to reduce > production costs. At the moment the desktop housing (excluding the top) > is made from a rectangle sheet of steel. The unit can have a maximum > width of 512mm and a maximum depth of 662mm. In order for the internal > components to cool properly, there has to be a minimum of 0.0144 cubic > metres of air contained within the box. Calculate the height of the box > such that the components are able to cool. Any help would be great?! Reading the question literally, I agree with WE's answer of about 24.33 mm, which gives the minimum surface area for a volume of 0.0144 m^3. (It would be better if the question specifically asked to minimise area, but I suppose that's what we're meant to understand by reduce production costs.) Nevertheless, this answer seems trivial and unsatisfactory. The 512 mm and 662 mm constraints do not come into play - they could be anything, provided greater than 24.33 mm. I also don't understand the significance of the word minimum in minimum of 0.0144 cubic metres of air. Why not just there has to be 0.0144 cubic metres of air contained within the box? I find it hard to believe that 24.33 mm really is the intended answer. I think the question is actually asking for something else, though I've no idea what. Either it is very badly worded or some essential information has been omitted in the transcription. Or maybe I am just stupid! > presented with this Question at Uni and I could do with some help on how > to go about solving it: A company that manufactures computer housings is trying to reduce > production costs. At the moment the desktop housing (excluding the top) > is made from a rectangle sheet of steel. The unit can have a maximum > width of 512mm and a maximum depth of 662mm. In order for the internal > components to cool properly, there has to be a minimum of 0.0144 cubic > metres of air contained within the box. Calculate the height of the box > such that the components are able to cool. Any help would be great?! Reading the question literally, I agree with WE's answer of about 24.33 > mm, which gives the minimum surface area for a volume of 0.0144 m^3. > (It would be better if the question specifically asked to minimise > area, but I suppose that's what we're meant to understand by reduce > production costs.) Nevertheless, this answer seems trivial and unsatisfactory. The 512 mm > and 662 mm constraints do not come into play - they could be anything, > provided greater than 24.33 mm. I also don't understand the > significance of the word minimum in minimum of 0.0144 cubic metres > of air. Why not just there has to be 0.0144 cubic metres of air > contained within the box? I find it hard to believe that 24.33 mm really is the intended answer. > I think the question is actually asking for something else, though I've > no idea what. Either it is very badly worded or some essential > information has been omitted in the transcription. Or maybe I am just > stupid! Sorry, finger trouble ... I meant 243.3mm I think. Also, having just read PS's response, maybe that's along the right lines. Maybe it is just that classic box-constructing calculus problem in heavy disguise? Ho-hum... too tired to think now! > presented with this Question at Uni and I could do with some help on how > to go about solving it: A company that manufactures computer housings is trying to reduce > production costs. At the moment the desktop housing (excluding the top) > is made from a rectangle sheet of steel. The unit can have a maximum > width of 512mm and a maximum depth of 662mm. In order for the internal > components to cool properly, there has to be a minimum of 0.0144 cubic > metres of air contained within the box. Calculate the height of the box > such that the components are able to cool. Any help would be great?! A strangely stated problem. I interpret it as follows (classic Calc III problem): An open top box is to be constructed from a rectangular sheet by cutting out squares from each corner of the sheet and folding up the sides. The box is to have a volume of 0.0144 m^3 and the width and depth are not to exceed 0.512 and 0.662 meters resp. What dimensions will minimize the area of the rectangle? Let z be the length of the sides of the squares to be cut out and let x and y be the width and depth. The rectangular area is A(x, y, z) = (x + 2*z)*(y + 2*z). There is a constraint of V(x, y, z) = x*y*z - 0.0144 = 0. Using the method of Lagrange, one easily gets y = x and z = x/sqrt(8) so x = y = cbrt(sqrt(8)*0.0144) =~ 0.344 meters (which is smaller than both 0.512 and 0.662) and z =~ 0.122. Now, what was the _real_ problem? -- Paul Sperry Columbia, SC (USA) > How would you prove that the image of a bounded set mapped by a bounded > linear transformation is also bounded? This follows from the definitions very easily. If you're stuck, tell us what is sticking you. it yet. Please solve it with work and email me the entire solution and work and I will PayPal you $5. Here's the problem... lim sin(3(x+h))-sin(3x) >h->0 ------------------- > h sin(3(x+h)) = sin(3x+3h) Expand sin(3x+3h) and it's cake. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ I feel a wave of morning sickness coming on, and I want to be standing on your mother's grave when it hits. 1/h [sin(3x+3h)-sin(3x)] = 1/h [sin(3x)cos(3h)+sin(3h)cos(3x)-sin(3x)] {trig identity} = 1/h [sin(3x)[cos(3h)-1]+sin(3h)cos(3x)] = sin(3x) 1/h [cos(3h)-1] + cos(3x) 1/h sin(3h) Let k=3h... (1) = 3sin(3x) 1/k [cos(k)-1] + 3cos(3x) 1/k sin(k) Now, as h->0, so does k->0. To proceed, you need to know (2) lim_{k->0} 1/k sin(k) = 1 and (3) lim_{k->0} 1/k [cos(k)-1] = 0. Both of these are dervied at the end. Then, the first term in (1) is 0, and the second term in (1) is 3cos(3x). So, lim_{h->0} 1/h [sin(3x+3h)-sin(3x)] = 3cos(3x). Lastly, to prove those two identities (2) and (3): For the first limit (2), note that sin k <= k <= tan k. sin(k)/sin(k) <= k/sin(k) <= tan(k)/sin(k). 1 <= k/sin(k) <= 1/cos(k). 1 <= lim_{k->0} k/sin(k) <= lim_{k->0} 1/cos(k). 1 <= lim_{k->0} k/sin(k) <= 1. 1 <= lim_{k->0} sin(k)/k <= 1. So we conclude lim_{k->0} sin(k)/k = 1. For the second limit (3), note that [cos(k)-1]/k = [(cos(k)-1)(cos(k)+1)]/[k(cos(k)+1)] = [cos^2(k) - 1]/[k(cos(k)+1)] = [sin^2(k)]/[k(cos(k)+1)] = sin(k)/k sin(k)/[cos(k)+1]. As k->0, we already show that sin(k)/k -> 1, and sin(k)/[cos(k)+1] -> 0. So, we conclude lim_{k->0} [cos(k)-1]/k = 0. I hope I didn't make any mistake. And there's no need for the paypal thing. :) -kira > Hi. I'm working on an limit problem for my math class. No one has solved it yet. Please solve it with work and email me the entire solution and work and I will PayPal you $5. Here's the problem... lim sin(3(x+h))-sin(3x) > h->0 ------------------- > h >For what values of a there are exactly two real numbers that satify the equation: x^3 + (a-1)x^3 + (9-a)x - 9 = 0 ? -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ I feel a wave of morning sickness coming on, and I want to be standing on your mother's grave when it hits. >> 1. Solve for x: log (lower) 7, x + log (lower) 7 = log (lower) 7, root >> square (upper) 24 >Would you clarify what you mean by lower and upper? >So you mean subscript and superscript? Also, the second term in the OP's equation seems to be missing an operand. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ I feel a wave of morning sickness coming on, and I want to be standing on your mother's grave when it hits. >Could anyone help me solve these nine problems please? I need help! Here's some help: If you want to learn math, you need to do problems. Don't just post your homework: show us what you did on each problem. If you have no idea where to start on a problem, reread the most recent sections of your textbook and work through the problems with pencil and paper. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ I feel a wave of morning sickness coming on, and I want to be standing on your mother's grave when it hits.