If (a) is a positive constant, find the length of the curve between x=1 and x=2. For what value of (a) is th..
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The length of the curve between the two points according to my calculation was:\n\nd = (3/2a) + (a*ln(2)/4)\n\nThe length is minimum under the condition that\n\na = √(6/ln(2))\n\n\n\nI don't have the time to show the full calculations, but I'll give you some key points:\n\n1. Find the line element 'ds' using the formula:\n   ds = √(dx^2 + dy^2)\n   -or-\n   ds = dx * √(1 + (dy/dx)^2)\n\n2. Evaluate (dy/dx) and substitute in above equation. \n\n3. Integrate the line element 'ds' determined above from x=1 to x=2 - this should give you the length of the curve in terms of 'a'\n\n4. To find the minimum length, differentiate the equation for length with respect to 'a', make the equation equal zero & solve for 'a'. This should give you the values for 'a' that is either for minimum or maximum. Now, differentiate the new equation once more & substitute the different values of 'a'. For each case, check if the equation yields a positive value - this means that it is a minimum case.
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Between any two points there exists a line, and between three points you can find a parabola.\n\nSo, you have a parabola of the form:\nc*(x-a)*(x-b)\nSince 4,0 is a point, you can let b = 4\nso the parabola is c*(x-a)*(x-4)\nsubbing in the other points, we get:\n-3 c + 3 c a = 3\n12 c + 6 c a = 1\nSolving gives\nc = -5/18\na = -13/5\nSo the curve is:\n-5/18*(x+13/5)*(x-4)\n=-5/18x^2 + 7/18 x + 26/9
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x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
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sidharth has solved the problem correctly and given you the two values for x that satisfy the conditions of the problem.\n\nHowever, he didn't specify the POINTS on the curve.\nThey are (0,0) and (-2,-2).
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x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
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It depends on what the word "mean" means. :-)\n\nRoughly speaking, under the specified conditions there is a point on the curve between x=a and x=b where the slope of the curve equals the slope of the secant from (a,f(a)) to (b,f(b)).
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The only interesting point in this curve will be the inflexion point which you find by doing the derivative twice and equating to zero.\n\nf'(x) = 3x^2 + 1\n\nf"(x) = 6x = 0 <===> x = 0\n\nf(0) = -1\n\nThe curve will have an inflexion point at (0,-1).\n\nf(0) < 0, f(1) > 0 then:\n\nThe curve will also have a single real root in [0,1] (x = 0.6823)
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if (1/x) + 2= (c/2)\n--------------------------------\n\n1/x = (c/2) - 2\n----------------------\nx = 1/((c/2) - 2)\n       (OR)\nx = ((c/2)-2)^-1\n-----------------------
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x=4\n\nplease solve...\n\n8-(1) :-) ;-(
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Just split it into parts - when is x-x^2<0, and when is it >0?\nThat bit has roots 0 and 1, so it is positive between 0 and 1, negative everywhere else.\n\nThus, ∫(x=-1 to x=0) (x^2-x) dx + ∫(x=0 to x=1) (x-x^2) dx + ∫(x=1 to x=3) (x^2 - x) dx\n\n= [(1/3 x^3 - 1/2 x^2) from -1 to 0] + [(1/2 x^2 - 1/3 x^3) from 0 to 1] + [(1/3 x^3 - 1/2 x^2) from 1 to 3]\n= (0- -5/6) + (1/6 - 0) + (9/2 - -1/6)\n= 17/3.
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Example: Find the length of a helix which winds three times around a cylinder of height 12" and radius 3"\n\nAnswer: divide the cylinder into three sections. The helix goes once around a cylinder of height 4" and radius 3"\n\nThe surface of the cylinder can be "peeled" into a rectangle 4" by 2 * pi * 3" = 18.850". The helix is a diagonal of that rectangle. So its length is sqt(4^2 + 18.850^2) = 19.27"\n\nIt goes around the cylinder three times and its total length is therefore 57.81"
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f(x) = x^m(1-x)^n\nm zeros at  x = 0 and n zeros at x = 1 \nSo x-axis is tangential to the curve at x = 0 and x = 1\n\nf'(x) = mx^(m - 1)(1 - x)^n + x^m * -n(1 - x)^(n - 1)\n= x^(m - 1) * (1-x)^(n - 1) * (m(1 - x) - nx)\n=  x^(m - 1) * (1-x)^(n - 1) * [m - (m + n)x]\n= 0 for stationary points.\n\nSo (m - 1) zeros at x = 0, (n - 1) zeros at x = 1 and a zero at x = m / (m + n)\n\nNow there has to be a turning point between x = 0 and x = 1 as f(x) is a polynomial and therefore continuous and differentiable for all x\n\nSo examine the possibilities\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... - .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... - .......... + ........... + ... Local minimum at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n ............... + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... + .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m .................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... + .......... + ........... + ... Local minimum at (1, 0)
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***************\nQUESTION 1\n***************\nOn the first question, we first have to find the equation that we will be solving for.  \n\nAssume that x is one integer.\nWe know that x+1 is the other integer because they are consecutive.\n\nSo we now put the wording into an equation\n1/3(x + (x+1)) = x - 5\n\nNow we solve the equation\n1/3(2x + 1) = x - 5\n2/3x + 1/3 = x - 5\n16/3 = 1/3x\n16 = x\n17 = x + 1\n\nOur two integers are 16 and 17.\n\nNow we check our answer.\n1/3 of 33 = 11 & 16 - 5 = 11 We are good! :)\n\n\n***************\nQUESTION 2\n***************\nEvery hour, Amy and Randy together walked 3.6 + 4.2 km apart from each other.\n\n3.6 + 4.2 = 7.8\n\nSo, every hour they will be 7.8 more km apart from each other.  We only want 10 minutes though, and there are 60 minutes in an hour. To get how far apart they will be in 10 minutes, we take:\n7.8 * 10/60 = 1.3\n\nRandy and Amy will be 1.3 km apart after 10 minutes.\n\nWe are done! :)
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Let P be (a,b). The slope of the tangent at P is found from the derivative of the curve: -128 / x³, giving the equation of the tangent as\n\n          y - b = ( -128 / x³ ) ( x - a )\n\nNow find where this tangent lines cuts the X and Y axes. Let x=0 and solve for y. Let y=0 and solve for x to get\n\n          ( 0 , b + 128 / a² ) and ( a + a³b / 128 , 0 )\n\n(a, b) is on the curve so b = 64 / a². Use this to simplify the two intercepts to\n\n          ( 0 , 192 / a² ) and ( 3/2 a , 0 )\n\nUse Pythagoras to find the length between these points as the square root of\n\n          f(a) = 192² / a^4 + 9 / 4 a²\n\nNow the problem is to minimize this expression (ignoring the square root, because it is an increasing function). This is a standard minimization calculus problem. Set f '(a) = 0, and check f "(a) > 0 to find that the minimum occurs when a = 4√2, so the required point is \n\n          ( 4√2 , 2 ).
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First find the length of the of z. Call the length L.\n\nL^2 = (sqrt(3))^2 + 1^2\nL^2 = 3 + 1 = 4\n\nL = 2.\n\nNow find the angle that x makes with the x axis. Call the angle t\n\ntan(t) = 1/sqrt(3)\nt = 30 degrees.\n\nSo in polar form, z = 2(angle sign)30 degrees
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A=pi*x*y \n\nwhere x is small radius from centre to circumference and y is large radius from centre to circumference.
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The correct answer is d.\n\nf(1-x) will give a negative number for all positive 'x' values.\n\nbut squaring these negative number will always give a positive number i.e (-4)^2 = +16.\n\nTherefore in the equation f(x) = x^2(1-x)^2;\n\nYou can see that for all positive or negative numbers you will have a positive value. Again,\n\nLet x=5; Then 1-x = -4;\n\nFor the equation d.. 1-(-4) = 5. i.e the original number.\n\nHence, f(5) = 5^2 (-4)^2 = 25*16;\nAlso, f(-4) = (-4)^2 (5)^2 = 25*16;\n\nThus, for all values of x... Equation d is TRUE.