Find a point on the curve...?
925420
I don't know the right english word, but I try.\nFind a line that touch the curve y=x^2 at (x1,y1) which (x1,y1) lies on the curve.\nFind another line that perpendicular with the first line that across (x1,y1) and (18,0)\n(x1,y1) is the closest point on curve to (18,0)\n\nthe common formula for second line is y=ax+b\nsubstitute (18,0) to this line we get  0=a*18+b\nwe get b=-18a\nnow we find (x1,y1) by combining curve y=x^2 with line y=ax+b\nsubstitute y we get   x^2=ax+b\nx^2-ax+18a=0  ..... (formula 1)\nsince line y=ax+b perpendicular with line touch curve at (x1,y1) then a=- 1/(dy/dx) = -1 / (2x1)\nsubstitute a in formula 1 we get  X1^2 - (-1/2x1)x1 +18(-1/2x1)=0\nmultiply the new formula with 2x1 we get\n2x1^3+x1-18=0\nwe get x1=2\nthen y1=x1^2\nwe get y1=4\nthe answer is (2,4)\nHope this satisfy you
950608
get some graph paper. plug in numbers for x and calculate its equivalent y value. do that 5 times, and you should have your curved line. If not, use higher or lower numbers than you used the first time ( I like starting with 0) Then  draw a line perpendicular to the most bent part (aka axis of symmetry, aka cut the line in two so both sides of the new line look like mirror images), find 2 points on that cutting line, and there you go.
1106036
Rise over run, or (y2-y1)/(x2-x1).  If your first point was (x1,y1) and your second point (x2,y2), then it's easy:\n\n[8 - (-4)] / [14 - 10] = (8+4)/(14-10) = 12/4 = 3\n\nThe slope of the line is 3.
755689
You want to minimize the length of the line segment running through points (0,A), (8,15), and (B,0).\n\nthis line is represented by the equation:  y = (15/(8-B))x + A\nsolve for A:\nA = y - (15/(8-B))x\nplug in values for x and y using the one known point (8, 15):\nA = 15 - (120/(8-B))\n\nlet f be the length function you want to minimize.\nf(B) = sqrt(B^2 + (15- (120/(8-B)))^2)\n\nnote that you are only interested in values of B>8.\n\nnow you can take the derivative of f using the chain rule and then solve for f'(B) = 0...\nor you can use a graphing calculator and find the minimum on the interval (8, inf).\n\nB should come to 20.164 or so.\nplug this value into f:\nf(20.164) = 32.013, approximately.
924809
You can take the deriviative of a function (say f ' (x)) and substitute for x the x-coordinate of a point somewhere on f(x) to get the slope of the tangent line that passes through that point on the curve.\n\nThe tangent line is the best approximation of the curve it is tangent to at the point of tangency.\n\nAlso, you can use point-slope form ( y-yo = m (x-xo) ) to find the equation of the tangent line.
900462
It really depends on how the equation of the path/curve is originally specified.\n\nIf it's specified in terms of x and y, an easy parameterization is:\n\nx = t\ny = f(x) = f(t)\n\nBut there are many other ways to do it.
180609
I would differentiate the function of the curve (I assume you know about differentiation) and then find the value of gradient at the point you want... Example\ncurve y=x^2 when x=1\nso dy/dx=2x and you want the gradient when x=1, so we get gradient=2. \nNow I would write this as a 3 dimensional directional vector. If a line has a gradient of 2, this means that for every 1 step in the x direction you get 2 in the y direction and 0 in the z direction so we get the directional vector (1,2,0) and for the x axis (the ground you are measuring the angle from) you only have 1 step in the x direction so the directional vector is (1,0,0). Now we have 2 direction vectors and we want to find the angle between them.  This can be done using a theroem called the dot product where a.b=mod(a)*mod(b)*cos(theta) where theta is the angle you want... a.b is the dot product and what you do is you times the x values for both vectors and add the y values multiplied together and add the z values muliplied together so with a=(1,2,0) and b=(1,0,0) we get 1+0+0=1 and mod(a) means the size of (1,2,0), its like 3D pythagorus so you get sqrt(1+4)=sqrt(5) and for mod(b) we get sqrt(1)=1 so we end up with 1=sqrt(5)*cos(theta), rearrange the equation and we get theta=cos^-1(1/sqrt(5))=63.4 degress, hope this helps
819957
Start with a graph of:\nf(x) = x\n\nThat's a straight line going through the origin with a slope of 1.  It should look like a 45 degree angle if you draw it correctly.\n\nNow to transform this to:\nf(x) = | x |\n\nYou have to think of each value smaller than 0.  Rather than -1, for example, you'll have 1.  Rather than -2, you'll have 2, etc.  This is like "mirroring" the graph that is below the x-axis so that it is above the x-axis.\n\nThis results in a "V" shape.  A slope of -1 down to the origin, then a slope of 1 from the origin.\n\nFinally:\nf(x) = | x | - 2\n\nFor every value that results, you now subtract 2.  For example, x = 0, gives -2, rather than 0.  This is true for every point x, making every point y smaller by 2.  It would look like you shifted the whole graph down 2 units.\n\nSo it would be "V" like before, but it would go down at a slope of -1, cross the x-axis at -2, and continue to (0, -2).  Then it would slope up (m=1) through the x-axis at 2 and continue upward.\n\nYou can go to the first link and type 'abs(x) - 2' in the box.  The resulting graph is shown as the 2nd link.
442603
A curve that is rising..................... I guess.
65888
a parabola facing up!\n\nwith center (0,1)
1095247
For a fixed a and n, and allowing b and c to vary, the curve would be a hyperbolar centered at the origin. You will only be taking the rational points on that curve, though.
303208
You are referring to locus/parametrics. The shape you are referring to in relation of a locus of a point and a line, is a PARABOLA. It looks like an curve/hill kinda thing.
612056
the local maxima can be found at the points where dy/dx is equal to zero and (d^2.y/(dt)^2) is less than zero. The local minima can be found at the points where dy/dx is equal to zero and (d^2.y/(dt)^2)(read as dee sqared y by dee tee squared) is greater than zero.\n\nThe explanation is that if u were to draw the graph of watever function u want to find the maxima and minima of,.... at the point where its max, the tangent is zero and the differential of the tangent is sloping downward!! And it slopes upward for the point of minima....\n\nHope this helps.\n\nYou better gimme best answer for this.
271844
If you need the tangent, take the derivative of the function at the specific point, which will give you the slope. Since you have taken the derivative at some specific point (a,f(a)) you also have a point. Now use the point-slope form to figure out the equation. Then simply get a graph paper and graph it.\n\nExample: f(x)=3tan(2x)\n\nf'(x)=6[sec(2x)]^2 (I assume you can take derivatives.)\n\nYou have a point (a,f(a))\n\nNow use the point-slope form: y-f(a)=f'(x)(x-a)\ny-3tan(2a)=6[sec(2a)]^2*(x-a) is the answer in this case. \n\nOr in function form: y=6[sec(2a)]^2*(x-a)+3tan(2a)\n\nTo graph it, find 2 points on the equation, then draw a straight line between them. That is your tangent.
41570
A parabola with its base at 0,1.  Go to http://www.coolmath.com/graphit/ Enter x^2+1 and see it for yourself!  :)
201825
Well um...\n\na = r''\n\na = (tangential acceleration * unit tangent) + (normal acceleration * unit normal) = (at * T) + (an * N)\n\nT and N are your unit tangent and unit normal of the function that defines your position.\n\nat = (r' dotted with r'') / (|r'|)\n\nThose are the tangential acceleration components but for this problem i think that you are going to have to reparamiterize your curve.
134706
I'd plot the points and figure out the slope (vertical change/horizontal change).  In this case, it would be -1 [2 / -2].  Then trace out the line to see at what point it crosses the Y axis.  In this case, it would be -2.  Then plus those numbers in the the equation "Y = (slope)X + (intersection of Y axis)"  So, if I'm not mistaken, the solution should be "Y = -X - 2".