Find the coordinates of the point P on the curve where the segment of the tangent line at P that is cut off by
912852
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 ).
187139
Yes (-1,0) and (1,-2) share the tangent line y=-x-1\n\nwhich y=-x-1 is tangent to the function y= f(x)=x^4-2x^2-x\n\nNote that (-1,0) and (1,-2) satisfy y=x^4-2x^2-x
759289
First you'll need to find dy/dx, which is the first derivative of curve C.\n\ny=4x² + (5-x)/x\n\ndy/dx = 8x -5/x²\n\nNow we need to find out what dy/dx equals at point P. The x-coordinate of P is 1, so we plug in 1 for x and solve:\n\ndy/dx= 8(1) -5/(1)² = 8 - 5 = 3
697906
let y = arc tan (x)\n\ntan y = x\ndifferentiate\n\nsec^2 y dy/dx = 1\n\nor(1+tan ^2y) dy/dx = 1\n\ndy/dx = 1/(1+tan^2 y) = 1/(1+x^2)
706522
It's often handy to draw yourself out a little sketch.  From it you can see the location of the given point.\n\nTo identify the angle the given point makes with the x-axis and the y-axis use trigonometry.  Switch to radians.\n tan Θ = opp./ adj.\n Θ = tan^-1 opp./ adj.\n Θ = tan^-1 (-0∙4425./ -0∙8968)\n Θ = tan^-1 (-0∙493 210 52)\n Θ =  0∙458 370 613 rad.\n    This is the value of the angle below the x-axis.  You need to include the angle above the x-axis, which is 180º or π radians.\n Θ =  0∙458 370 613 rad. + π rad.\n Θ =  (0∙458 370 613 + 3∙14) rad.\n Θ =  3∙598 370 613 rad.\n Θ ≈  3∙6 rad.
506648
use a protractor - thats' the semi circle type thing...\n\nmeasure the angle on a straight line which is 180 degrees in total, the obtuse (larger) angle and the acute (smaller) angle will equal 180 degress so calc the obtuse one and subtract this by 180 leaving the acute angle.
735905
a). If it touches the x-axis, then y is zero.  \nSo x^2 + 0 + 0 = 7\n\nx^2 = 7\n\nx = sqrt 7 or - sqrt 7  \n\nThe points are (sqrt 7, 0) and (-sqrt 7, 0)\n\nb).  Differentiate implicitly like this:\n\nd/dx x^2 + d/dx xy + d/dx y^2 = d/dx 7\n\n2x + (x dy/dx + y) + 2y dy/dx = 0\n\nx dy/dx + 2y dy/dx = -2x - y\n\ndy/dx (x+2y) = -2x - y\n\ndy/dx = (-2x-y)/(x+2y)\n\nc,d).  Just plug in the coordinates from (1)
152408
Use excel to give delta y/delta x on the curve. Look for a point where this value (the derivative) shoots up in value. That's your point.
963751
find the first derivative for the curve, and then substitute the x coordinate for the slope at that point.\n\nfor example, for the graph y = x^2, the first derivative is 2x. to find the slope at where x = 1, sub x = 1 into 2x and the slope at that point is 2.\n\nalternatively, draw a tangent to the point where u want to find the point slope and the slope of the tangent is the point slope. note that this method's accuracy depends on how accurately you draw your tangent
534712
Actually sometimes. A secant passes through a circle at two points, a tangent at one. You COULD have a secant hit the tangent at a point other than the point of tangency, and there is indeed a theorem around for determining the relative lengths of lines and so forth, but many questions dealing with area and arc length put the vertex at the point of tangency for simplicity's sake.
488924
since the tangent is parallel to line 8x=9y... the relation 8x=9y holds good for the tangent so\nx=9/8 y\nsince the tangent touches the ellipse the points must be on the ellipse so substitute x=9/8 y in the equation of the ellipse 4x^2 + 9y^2 = 1\n\n4(9/8 y)^2+ 9y^2 = 1\ny^2(81/16 + 9) =1\nY^2(225/16)=1\ny^2=16/225\ny=+(or)- 4/15\ntherefore \nx= 9/8 (+(or)- 4/15)\nx=+(or)- 3/10\ntherefore the points are (4/15,3/10) and another tangent parallel to the previous on at (-4/15,-3/10)
833915
Let the center of the inner circle be O and the center of the 8" circle be P.  The inner circle is tangent to the two 8" sides of the sector at A and B, and also tangent to the circumference of the 8" circle at C.\n\nDraw the line PC, which, by symmetry, passes through O, bisecting the 96-degree angle APB.\n\nAngle OAP is a right angle due to tangency.  Let PO be x, and we have OC = r.\n\nThen sin OPA = sin 48 = r/x, and x = r/sin 48.\n\nPO + OC = x + r = r/sin 48 + r = r(1 + 1/sin 48) = 8\n\nr = 8/(1 + 1/sin 48) = 3.41 inches  (Answer)
868167
In polar coordinates area is\n\nA = ½∫r² dθ\n\nThe area of the sector bounded by the curve r = f(θ) and the two radial lines defined by θ1 and θ2 is\n\n½∫[θ = θ1 to θ2] f(θ)² dθ\n\nIn your problem f(θ) = 2 cos 3θ. The curve starts when θ = 0 at (2,0) and continues. It first meets (0,0) when 2 cos 3θ = 0, or θ = π/6 making a region of half a loop.\n\nSo, area of one loop =\n\n2 × ½∫[θ = 0 to π/6] ( 2 cos 3θ )² dθ\n\n= 4∫[θ = 0 to π/6] cos² 3θ dθ\n\n= 4∫[θ = 0 to π/6] ½ ( 1 + cos 6θ ) dθ\n\n= 2 { ∫[θ = 0 to π/6]  dθ + ∫[θ = 0 to π/6] cos 6θ dθ }\n\n= 2 { [θ = 0 to π/6: θ]  + 0 } because the area over the range 0 to π/6 of cos 6θ cancels to 0\n\n= π / 3
986326
when x is 0, find y. plot the point (0, <<whatever you got>>).\nwhen y is 0, find x. plot the point (<<whatever you got>>, 0).\nConnect those two points, and extend the line.
95382
Taken from about.com:\nTo find the midpoint of a line section the formula is: \n\nx = (x1 + x2)/2 \nand \ny = (y1 + y2)/2 \n\nwhere x and y are the coordinates of the midpoint, and (x1,y1) are the coordinates of the first point and (x2,y2) are the coordinates of the second point. Let's plug it in to the equation. \n\nx = (7+10)/2 = 17/2 = 8.5 \ny = (3+6)/2 = 9/2 = 4.5 \n\nSo the midpoint is (8.5,4.5)
510334
x = a*sin(ang)\ny = b*cos(ang)\n\nang goes from 0 to 360 degress, a and b determine how oval it is (if a=b it is a circle)
292936
You cut at 1/4 of the distance from the center to the edge.  If you want it more exact, it's 0.2649 of the radius.  Steve is wrong, size of the cookie doesn't matter.\n\nIf you want the trig: the cut is a chord parallel to the horizontal diameter.  Draw radii to the ends of the chord, and an altitude from the center of the circle to the (center) of the chord.  For convenience let radius r=1.  Length of altitude = d.  Half the length of the chord = w.  And angle between altitude and radius = x (in radians). \n\nNow area of the sector is 2x/2pi*pi*r^2 = (so nice!) just x.  Area of the triangle is 1/2 base * height = d*w.  And we eventually want the value of d, distance from center to the cut.\n\nSo area of the slice = area of sector - area of triangle=\nx-dw=1/3*area of whole circle = 1/3*pi*r^2=pi/3\nSince d=cos(x) and w=sin(x), \nwe need to solve x-cos(x)*sin(x)=pi/3.\nUsing a graphing calculator, we get x=1.302662837 radians, for which cos(x)=d=0.2649320846.