This is a example from my Calc book and I'm stuck on some parts....here goes the question and the solution.\n\nQuestion:\nFind the length of the arc of the semicubical parabola y^2=x^3 between the points(1,1) and (4,8).\n\nSolution:\nFor the top half of the curve we have y=x^3/2 (dy/dx)=(3/2)x^1/2\nand so the arc length formula gives\nL=integral [1,4] sqrt(1+(dy/dx)^2) dx = integral[1,4]sqrt(1+(9/4)x)dx\nIf we substitute u=1+(9x/4), then du=(9dx)/4. When x=1, u=13/4; when x=4, u =10. Therefore (and this is the part where I am lost I understand everything up to this point):\nL=4/9integral[13/4, 10] sqrt(u)du=4/9*(2/3)u^3/2]integral[13/2,10]\n=(8/27)[10^(3/2)-(13/4)^(3/2)]\n=1/27(80sqrt(10)-13sqrt(13))\nokay so how did they do this please help...is there an easier way to do this??\n\nAnd for more info:\nIf we use Leibniz notation for derivatives, we can write the arc length formula as follows:\nL=integral[a,b]sqrt(1+(dy/dx)^2)dx
125216
If you understand everything up through the u-substitution then you have done the tough part. If the original integral was:\n\nintegral[1,4]sqrt(1+(9/4)x)dx\n\nthen letting u = (1+(9/4)x)  and du = 9/4 dx *and* substituting the new u-values for the bounds (13/4 and 10 in place of 1 and 4).....\n\nThe new integral becomes:\n\nintegral[13/4, 10] sqrt(u) dx ....but dx = 4/9 du (see above) so now:\n\nintegral[13/4, 10] (4/9) sqrt(u) du  - I'm sure you are comfortable moving the (4/9) outside the integral to get:\n\n(4/9) integral[13/4, 10] sqrt(u) du. Rewrite sqrt(u) = u^1/2 to get:\n\n(4/9) integral[13/4, 10] u^1/2 du and integrate...\n\n(4/9) (2/3 u^3/2) from [13/4, 10]....better as\n\n(8/27)(u^3/2) from [13/4, 10] - now substitute to get:\n\n(8/27)(10^3/2 - (13/4)^3/2)\n\nIf 10^3/2 = 10sqrt(10) and 13/4^3/2 = 13sqrt(13)/8....then you have\n\n(8/27)(10sqrt(10) - 13sqrt(13)/8)\n\nDistribute the "8" from the 8/27 to get your final answer:\n\n(1/27)(80sqrt(10)-13sqrt(13))\n\nLet me know if that makes sense.
461506
∫tan^(1/3)x dx\nI can predict that this'll involve the use of partial fractions.\n\nLet tanx=u^3\n\n=>(sec x)^2 dx=3u^2 du\n\nnow, (sec x)^2=1+ (tan x)^2=1+u^6\n\n=>dx=(3u^2)/1+ u^6\n\ntherefore, ∫tan^(1/3)x dx=3∫[u X u^2/(1+u^6)]du\n\n=>3∫[u^3/(1+u^6)]du\nlet u^2=t\n=>(2u) du=dt\n\n=>3∫[u^3/(1+u^6)]du=(3/2)∫[t/(1+t^3)]dt\n\nNow, partial fractions will be used\nwe know that, t/(1+t^3)=t/(1+t)(1+t^2-t)...........since a^3 + b^3=(a+b)(a^2 + b^2 + ab)\n\nlet, t/(1+t)(1+t^2-t)=A/(1+t) + (Bt + C)/(1+t^2-t)\n\n=>t/(1+t)(1+t^2-t)=[A(1+t^2 -t) + (Bt + C)(1+t)]/(1+t)(1+t^2 -t)................taking LCM\n\nequating coefficients of u, u^2, constants on both sides\n=>1=-A+B+C...................equating coefficients of u\n     0= A+B .......................equating coefficients of u^2\n     0= A+C .......................equating coefficients of constant\n=>A=-1/3, B=1/3, C=1/3\n\n=> (3/2)∫[t/(1+t^3)]dt=\n(3/2)∫[(1/3)/(1+t)]dt + (3/2)∫[(1/3)(1+t)/(1+t^2-t)]dt\n\n\n=>(-1/2)log|1+t| +  (1/2∫[(1+t)/(1+t^2-t)]dt\n\n=>(-1/2)log|1+t| +  (1/4)∫[(2+2t)/(1+t^2-t)]dt \n\nwe know, d/dt(1+ t^2 - t)=2t -1\n=>(-1/2)log|1+t| +  (1/4∫[(2+2t+1-1)/(1+t^2-t)]dt \n\n=>(-1/2)log|1+t| +  (1/4)log|1+t^2 -t| + (3/4)∫1/(1+t^2-t)dt\n\nwe know, 1+t^2-t=(t-1/2)^2 +3/4\n\n=>(-1/2)log|1+t| +  (1/4)log|1+t^2 -t| + (3/4)∫1/[(t-1/2)^2 + 3/4]dt\n\n=>(-1/2)log|1+t| +  (1/4)log|1+t^2 -t| + \n(3/4)1/(2/sqrt(3) arctan[2(t-1/2)/sqrt(3)\n\n\nsubstituting original values t=u^2\n\n=>(-1/2)log|1+u^2| +  (1/4)log|1+u^4 -u^2| + \n(3/4)1/(2/sqrt(3) arctan[2(u^2-1/2)/sqrt(3)\n\nsubstituting original values u=(tanx)^1/3\n\n=>(-1/2)log|1+(tanx)^2/3| +  (1/4)log|1+(tanx)^4/3 -(tanx)^2/3| + \n(3/4)1/(2/sqrt(3) arctan[2((tanx)^2/3 - 1/2)/sqrt(3) + C\n\nwhere arctan(x) means tan inverse x\nresults used= \n1)∫1/(1+x)dx= log|1+x| + c\n2)∫1/(a^2+x^2)dx=\n(1/a)arctan(x/a) + C...........'a' is a constant\n\n\nP.S. Extremely sorry for the length of the solution but it is correct and each and every explanation is given.
600477
Calculus tells us a lot about optimizing functions. However, it isn't needed here. No matter what planet you're on, the height-vs-time graph of a free-falling object is always a parabola.\n\nThere are a lot of facinating facts about parabolas and conic sections in general, but let's get to answering the questions that you asked. ;)\n\nThe maximum or minimum of a parabola occurs at its vertex. The value of the independent variable ( t ) at the vertex of a parabola is -B/2A. Therefore, t = -16/2(-4) = -16/-8 = 2.\n\nA) The bullet reaches its highest point two seconds after it is fired.\n\nMoving on, the height of the bullet it determined by the formula itself: h = -4t^2 + 16t + 5. If the bullet is on the ground, then h = 0. So:\n\n-4t^2 + 16t + 5 = 0\n4t^2 - 16t - 5 = 0 [Multiply by -1 to work with leading positive.]\nt = [16 +/- sqrt(16^2 - 4(4)(-5))]/(2*4) [Quadratic Formula]\nt = [16 +/- sqrt(336)]/8\n\nThe negative root does no good, as the bullet cannot land before it was fired. But the positive root show us that:\n\nB) The bullet hits the ground approximately 4.3 seconds after is was fired.\n\nLast is actually the easiest part, although most students miss it. You know the bullet reaches its highest point when t = 2 from part A. What will be its height at that point? You already have the equation. h = -4t^2 + 16t + 5.\n\nh = -4(2)^2 + 16(2) + 5 = 21\n\nC) The bullet will reach a height of 21 feet.\n\nSo, to sum up:\n\nA) The bullet reaches its highest point two seconds after it is fired.\nB) The bullet hits the ground approximately 4.3 seconds after is was fired.\nC) The bullet will reach a height of 21 feet.\n\nI hope this helps! =]
435381
O.K., this is a quadratic equation (quadratic means that the highest power involved is a "squared").  \n\nUnfortunately, once I hit "preview", the drawing I tried to do of a quadratic loses all its shape. The closest I can come using a single character is U. Or maybe think of a U crossed with a V.\n\nHave you done any differential calculus? If you have, you'd know dy/dx = 4x + 3. Basically, this tells you how steep the curve is for any given value of x. (1) So, if you increase the co-efficient of x^2, you get a steeper graph. \n\n(2) If you increase the value of the co-efficient of x, this moves the lowest point of the graph to the left (thinking of the usual x-y axes), that is it decreases the value at which the graph attains its minimum value.\n\n(3) If you increase the value of the constant (in this case 4), then you increase the lowest value that the graph attains. In this case, with the numbers you have, the least value the graph attains is 4. If you set the constant to zero, then the graph touches the x-axis, if you set the constant to less than zero, then the graph crosses the x-axis in two places, those two places being given by\n\nx = [-b +or- sqrt (b^2 - 4a.c)]/2*a, where, in this case, a=2, b=3 and c = 4.
294447
alright... \n\nso y=2x²-6x+1?\n\nthere could be 2 things coming out of this:a parabola or an absolute value, but i think the parabola's what you need.\n\nthe equation for the parabola is y=a(x-h)+k\nyou need to do the method 'completing the square'\nbut first factor out the 2 from the first 2 terms so it becomes y=2(x²-3x+?)\nto make a perfect square, take half of the third term, square that to make the third term. so half of 3 is 3/2, squared is 9/4.\nbut you can't just put a 9/4 because it was never originally there. so you have to either subtract it back out or add it to both sides.\nso it becomes y=2(x²-3x+9/4-9/4)+1\nsee, you add the 9/4 and you subtract it. the +1 was the number you originally had. then you factor the first 3 terms inside the parentheses\nso it would become y=2(x-3/2)²+1\n                   y=a(x-h)+k\nnow, the axis of symmetry would be (-3/2,1)\n to answer the second question, i dont know what the equation is\n\nfor the third question you have to plug in the numbers\n(x+3)(x-5) I'm not sure what roots are because its different for me but I'm thinking its the answer\n"what makes this equal 0?" is what you have to ask. so if u want -3. you want positive 3, and vise-versa for 5\nthen you FOIL method it, which equals x²-2x-15
417601
This is a tricky one. You have to use a couple of steps to get this one. First you use a u-substituion to "simplify" it and then use a trig substitution.\n\nIf you let u=cos(x) then du=-sin(x) then the integral becomes\n&#92;int -1/(1+u^2) du.\n\nNow let u=tan(s) and du=sec^2(s) and the integral is\n&#92;int -sec^2(s)/(1+tan^2(s)) ds = &#92;int -sec^2(s)/sec^2(s) ds = &#92;int -1 ds.\n\nThe integral is -s + C where C is any constant. We now have to undo the steps to get it back in terms of x. First, we undo the trig substituion. Since u=tan(s) then the integral is\n\n-arctan(u)+C\n\nSince u=cos(x) we get\n\n-arctan(cos(x))+C
561823
IF you want to try and get it in that EXACT form dicatated by Mathematica, I suggest you start with u=1+sin(x) and then solve for x in terms of u and do some resubstitution.  You'll get the arcsin function, which becomes a polynomial in terms of another dummy variable when you start the first u*dv substitution.  Notice I said "FIRST", because to get it in the form that you want, you'll have to do it about 6-10 times.  Also, you'll have to do some "right triangle simplification" throughout the problem as well.  Of course, then you get to factor it.  Have at it.  I did this very thing, stopping at the 3rd u*dv substitution because it is just ridiculous.  \nI did, however, get several of the terms.  I think if you decide to do this, you're going to need lots of paper.  I was doing it symbollically in a CAD program to keep things organized and such, and it was already 5 pages.  Oh, and not to mention that this "answer" given by Mathematica isn't even a closed form "answer".  It left it in terms of a polylogarithm, which is still an integral hidden in notation.  \n\nThe point is that the fuction (among others) is best integrated numerically because there is no useful closed form for this function.  This is what I did and it took me 2 minutes.  \n\nx^2/(1+sin(x))=x^2 - x^3 + x^4 - (5/6)x^5 + (2/3)x^6 - (61/120)x^7 + (17/45)x^8 - (277/1008)x^9 .................\n\nIntegrate termwise, which is fricking easy.  \n\n(1/3)x^3 - (1/4)x^4 + (1/5)x^5 - (5/36)x^6 + (2/21)x^7......\n\nNotice it is not undefined (like the original function) and the error term goes to 10^-7 after the first 5 terms.  I'd say that's as accurate as anyone would ever really need to be, and if not, they can always expand this function further.\n\nTell your uncle he's a bastard just for me.  He's teasing you.
441011
Let u = x², so du = 2xdx\n\nSo, ∫ (x/1+x^4) dx =\n\n1/2 ∫ (1/1+u²) du =\n\n1/2 arctan(u) + C = \n\n1/2 arctan(x²) + C\n\n~ ♥ ~
138225
First, I'll assume you mean that this is an ellipse (parenthesis are important):\n((y+2)^2)/25 + ((x-3)^2)/16 = 1\n\nTo graph by hand, start at the origin.  \n1)The center is shifted- to find it's location, set (y+2) = 0 and (x-3) = 0, and solve for x and y. The center is therefore at (-2,3). \n\n2)The half length of the axis (the lines bisecting the ellipse) are given by the denominators of the x and y terms.  To find the length of each axis, take the square root of each denominator.  The "radius" in the y (vertical) direction is sqrt(25) = 5 units long, and in the x direction it is sqrt(16) = 4 units long.  \n\n3)Starting at the center point of the ellipseee, (-2,3), count up half the length of the axis in the y direction, or 5 units, and put a point.  Do the same thing down, and then repeat with the x axis, counting left and right 4 units, and putting points on your graph.  Draw an ellips through these 4 points, and you're done! \n\nThe major axis is therefore the y axis, and is 10 units long, and then minor axis is the x axis, which is 8 units long.\n\nIf you want to graph this on a calculator, you must solve for y:\n((y+2)^2)/25 + ((x-3)^2)/16 -((x-3)^2)/16= 1 -((x-3)^2)/16\n\n(((y+2)^2)/25)*25 = (1 -((x-3)^2)/16)*25\n\n((y+2)^2)^(1/2) = ((1 -((x-3)^2)/16)*25)^(1/2)\n\nNote that raising to the (1/2) power is the same as taking the square root. When you take the square root of a real number, your answer could be negative or positive, so you have 2 solutions, one for the bottom of the ellips, and one for the top:\n\n(y+2) = (((1 -((x-3)^2)/16)*25)^(1/2))\nand\n(y+2) = -(((1 -((x-3)^2)/16)*25)^(1/2))\n\nThe final equations you'd enter into your calculator are:\n\ny = (((1 -((x-3)^2)/16)*25)^(1/2))-2\nand\ny = -(((1 -((x-3)^2)/16)*25)^(1/2))-2
447929
divide through\n\ny = [x^(3/2)]/3 - 2x^(1/2) -x^(-1/2)\n\nthen take the derivative\n\ndy/dx = [x^(1/2)]/2 - x^(-1/2) + [x^(-3/2)]/2\n\nthen put x^(3/2) in the denominator\n\ndy/dx = [1/x^(3/2)]*{ [x^2]/2 - x + 1/2 }\n\nwe take out the one half to find\n\ndy/dx = [1/2x^(3/2)] * [ x^2 - 2x + 1 ]\n\nwhich is the same as writing \n\ndy/dx = [1/2x^(3/2)] * (x - 1)^2\n\nor a = -1 and b = 2
280552
Well ... thought I had solved it, but then I got stuck!!!  Integration is not the best thing I am at!!!!\n\nUsing "$" as the integral sine.\n\nUse integration by parts: $udv = uv - $vdu\n\nThen split the orignal question into:\n\ny = $ (2x-3)(1/(√(9-x^2))) dx\n\nu = 2x - 3\ndu = 2\n\ndv = 1 / √(9 - x^2)\ndv = 1 / √(3^2 - x^2)          ==> standard integral\nv = sin^-1(x/3)     \n\nThen use the integration by parts equation above to try and solve it!  \n\ny = (2x-3)(sin^-1(x/3)) - $2 / sin(x/3)\n\nThen you have to solve that integral by parts again (using "z" this time, so not to be confused with the whole equation which is equal to "y")\n\nz = $2 / sin(x/3)\n\nu = 2\ndu = 0\n\ndv = 1 / sin(x/3)\nv = 3[x / 3sin(x/3) + √(1 - (x/3)^2)]\n\nz = $2 / sin(x/3) = 6[x / 3sin(x/3) + √(1 - (x/3)^2)]\n\nOK, (I hope this is right) so ... then put that back into the above equation that I got after the first round of integrating by parts!\n\nThat is: y = (2x-3)(sin^-1(x/3)) - $2 / sin(x/3)\n\ny = (2x-3)(sin^-1(x/3)) - 6[x / 3sin(x/3) + √(1 - (x/3)^2)]\n\nWhich eventually simplifies down to:\n\ny = [(4x - 3)/(sin(x/3)] - 6√(1 - (x/3)^2)]\nor\ny = [(4x - 3)/(sin(x/3)] - √(36 - 4x^2)]\n\nLet me know if that was right, if you ever find out!!!!\n\nI hope it was ... good luck
426192
f(x) = x^2 / (x - 2 ) ^2?\n\n    (a) Find the vertical and horizontal asymtotes\nx=2\n    (b) FInd the intervals of increase and decrease\nfind f'(x) = ( (x - 2 ) ^2 2x - x^2 2(x-2) )/(x - 2 ) ^4\n= ( x-2) ( (x-2)2x-2x^2) / (x - 2 ) ^4\n= ( 2x^2 -4x -2x^2 ) / (x - 2 ) ^3\n= -4x/ (x - 2 ) ^3  \n=0\nif -4x =0, i.e. x=0 one critical point.\n    (c) Find the local and maximum and minimum values\n    (d) Find the intervals of concavity and the inflection points
269850
1) Let x:the length , y:the width of the picture.Note:x,y>0\nThe first equation we have is : x=y+4 (1)\nThe combined area has the length is x+4,the width is y+4       \nso the second equation we have is: (x+4)*(y+4)=140 (2)\nReplace (1) in (2)-->(x+4)x=140--> x^2+4x-140=0 -->x=10 or x=-14 .since x>0 then x=10(cm)-->y=6(cm).\n2)  Let x: the length,y:the width.Note:x,y>0\n    The first equation we have is x=y+3 (1)\n    The second is: the area=x*y=550 (2)\n    Replace (1) in (2)--> (y+3)y=550--> y^2+3Y-550=0 --> y=22 or y=-25.Since y>0 then y=22(cm)-->x=25(cm)\n3)  Let x,y:two positive integers.Supposed that x>y>0 \n    we have two equation:x*y=240(1) and x=y+8(2)\n    Replace (2) in (1)-->y(y+8)=240--> y^2+8y-240=0 -->y=12 or y=-20.Since y>0 then y=12-->x=20.
544885
For a), we are asked to assume this is a parabola, even though on a real bridge the cable does not assume such a shape.\n\nTo solve for k, put in\n\n143 = k * 640^2\nk = 143 / 640^2\n\nDo the computation yourself to at least one decimal place.\n\nFor b), we need to be able to compute the length of a section of a parabola. For that, we need the derivative of the curve,\n\nf'(x) = 2x\n\nThe formula for the length is the integral of \n\nsqrt(1 + 4x^2) \n\nover the desired range. To try to integrate this, we can use trigonometric substitution, defining an angle theta such that \n\n2x = tan(theta)\nx = tan(theta) / 2\n\nNow we take dx/d(theta) = sec^2(theta) / 2\n\nsqrt(1 + 4x^2) = sec^2(theta) / 2
164555
Let u = 1 + x^4 and then apply the Chain Rule. This allows your expression to be written as\n\n1/sqrt(u)^(1/2)\n\nwhich is equivalent to\n\n1/u^(1/4)\n\nwhich, in turn, is equivalent to\n\nu^(-1/4).\n\nDifferentiating this expression with respect to x yields the following:\n\n(-1/4)*u^(-1/4-1)*(du/dx).\n\nNote: du/dx = 4x^3.\n\nTherefore, the desired derivative is \n\n(-1/4)*(1+x^4)^(-5/4)*(4x^3).\n\nIf we simplify this a bit we are given\n\n-x^3/(1+x^4)^(5/4).
180664
Find the intersection points for the two curves.\n6x-x^2 = x^2 - 2x\n0 = 8x - 2x^2\n0 = x(8-2x)\nx = 0, 4\n\nIntegrate both functions\nfor 1st func.-->  y = 3x^2 - (x^3)/3\nfor 2nd func.-->  y = (x^3)/3 - x^2\n\nEvaluate integrals from x=0 to x=4 (intersection points)\nfor 1st integral--> A1 = 80/3\nfor 2nd integral--> A2 = 16/3\n\nTo find area of region bounded by both:\nA = A1 - A2\nA = 80/3 - 16/3\nA = 64/3\n\nThere's your answer!
245688
Good that you care to ask. The answer to this should be evident to you by integration (hope you have been introduced to calculus? If not, you will be this year.). \n\nI assume that you don't have much of a background.\n \nOK, let the cone be such that its height is H, base radius R. \n\nWhat you need to do is consider an infinitesimally thin section of the cone (a diagram should be so helpful here, unfortunately not possible) at height h from the bottom and of height (along the axis) 'dh', such that the radius of the section is a uniform 'r'. \n\n(Note that you have a uniform radius r to the section because you have chosen a very thin section of the cone.)\n\nNow, the surface area of the section considered is \n2*pi*r*dh*cosec x. (x is the cone half-angle)\n\nThe cosec term comes from basic trigonommetry, when you realise that the section is tilted and not exactly parallel to the axis. \n\nThe surface area  of the entire cone, indeed the curved surface area, is a summation of all such infinitesimal sections, each of surface area 2*pi*r*dh* cosec x \n(where r is the radius of each section). \n\nThis above sum can be found by \nintegrating 2*pi*r*dh*cosec x (that is a word in the language of calculus for finding out the sum of many small parts) over the entire height of the cone. \n\nalso realise that r = h tan x. \n\nso an integration of 2*pi*htan x*cosecx*dh gives us,\n\nCSA = 2*pi*sec x* (H^2/2)  \n    = pi*R* H (sec x/tan x)\n    = pi*R* H cosec x\n    = pi*R*L\n\nHope this satisfies.