For example the slope in cartesian plane is nothing but tanA where A is the angle made by a straight line with respect to X-axis. Slope at any poing on the curve is given by drawing a tangent at that point and the slope of the tangent gives the slope of the curve at that point. But in space(3d geometry) how can we draw a tangent to a curve and with respect to which axis?
202833
the answer is pie over 2x plus 7
391126
you have to look at the line a determine the slope. \nEx. look for a point where it is on the exact "cross shape" and start from 5 and go up the y-axes untill the point at the exact "cross shape" and the point where you stop on the y-axes" is parrallel to the x-axes. Now, go horizonally to the "exact-cross-shape point". Count how many you go up and how many you go across. Those counts will be your slope and if you go to the left then slope is negative and if you go to the right then the slope is positive.
528564
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.
533336
A curve has an horizontal tangent when its derivative equals 0.\nSo all you have to do is calculate the derivative of your function (which is y'= 6x^2 +6x -36) and solve the equation y' = 0.\nYou'll find two values of x, which are the two points where your curve has an horizontal tangent.
637895
in standard ntation we have:\n\nA=2\nB=sqrt[3]\nC=1\n\ntan2t = B/(C-A) = -sqrt[3]\nHence 2t=-60 deg\nt=-30 deg\n\nThe standard notation for a quadratic curve is:\nAx^2+Bxy+Cy^2+Dx+Ey+F=0\nA, B, C determine the shape of the curve.\nThey are affected by rotation.\nchanging D, E, F only changes the position of the curve in plane by displacement.
117615
For the benefit of all the folks who answered, question a), since (2,-1,0) is not ON the x-axis, a line through it can be parallel to the x-axis.  Equations of lines in 3-space are VECTOR equations with a simple form (kind of like slope and intercept): we need a point the line passes through -- (2,-1,0) -- and a direction to go -- parallel to the x-axis.  So any point on this line, P(x,y,z)= (2,-1,0)+m(1,0,0).\n\nFor b), you need dot product divided by product of magnitudes equal to cosine of angle.  So\n(x,2,4) dot (1,0,1)/[sqrt(x^2+2^2+4^2)*sqrt(1^2+0^2+1^2)=sqrt(2)/2\n(x+4)/[sqrt(x^2+20)*sqrt(2)=sqrt(2)/2....cross multiply\n2(x+4)=2*sqrt(x^2+20)\n(x+4)=sqrt(x^2+20)............square both sides\nx^2+8x+16=x^2+20\n8x=4\nx=1/2
421789
This is a two-part function.  Its graph consists of two rays (arrows).  On the left side of the graph you will have a ray (part of a line) that starts at -1 on the y-axis and goes up and to the left with a slope of -1.  On the right side is another ray that starts at 1 on the y-axis and goes up and to the right with a slope of 1.  The way you have the function written here both rays would have circles at the end, because the endpoints are NOT included in either.  There's a good chance that in the actual problem one or the other has the < or > symbol underlined to indicate "or equal to".  If it is, you'd fill in the circle to show the point on the y-axis is included.
486233
x=t^2 - 1\ny=t^4 - 4t\n\ntangent vector: ( 2t, 4t^3 -4)\nthe tangent vector is horizontal if:4t^3 -4 =0, so t^3=1, so t=1,\nin this case\nx= 0, and y = 1-4=-3\n\nthe tangent vector is vertical if: 2t=0, so t=0\nin this case\nx= -1 and y= 0\n`
268221
Let the coordinates of point P be (x,y). Here, x and y represent general points.\n\n1) Find an expression for the slope of OP. To help you get started, I will tell you the answer directly. It is (y/x).\n\n2) Find an expression for the slope of AP. Try this yourself!\n\n3) Use the statement "slope of OP is 1 less than the slope of AP" to construct an equation in terms of y and x.\n\n4) Carry out some algebraic manipulation to obtain your final answer!\n\n5) Lastly, take a look at your final answer. Are there some coordinates that point P cannot take? For instance, your final answer should be an equation that has (0,0) as 1 of the solutions. But can point P take this set of coordinates? What other coordinates can P not take?
384735
x=r(t-sint) and y= r(1-cost) are the parametric equations of the cycloid where r is the radius of the rolling circle and t is the angle through which the radius of the circle has turned from its starting point. The angle t must be measured in radians.\n\nFor simplicity let r =1 so that we have the unit circle.\nwith t=60 = pi/3 we get\nx = pi/3-sin(pi/3) = pi/3 -(sqrt 3)/2\ny = 1-cos (pi/3) = 1- 1/2 = 1/2\n\ndy/dx= sint/(1-cost)\n=[(sqrt 3)/2]/(1/2) = 3^0.5 (at the point where t= 60= pi/3)\n\nFrom this , you can find the slope at any point define by the parameter t.\n\nb) The tangent is horizontal when t= pi and is perpendicular at the cusp which occurs at 2pi and multiples thereof. The maximum value  is 2r (or 2 for the unit circle). The length of the base of the arch is 2pi *r or just 2pi for the unit circle.
261617
Draw a straight line.\nDraw a good cuve (mirror), using a compass or a protractor(for later mark a point as centre and place the protractor on this point coinciding with the axis.)\nThe curve should be of equal length on each side of principle axis.\nNow mark the pole, radius and focal point (twice the radius).\nIn case u use protractor, take focal lenght as twice the distance between the marked pont and pole.\n\nRemember how the rays reflect/refract, and accordingly draw two incident and two reflected/refracted rays.\nGive the standard sign convention...I hope plane mirror and prism are easier.
429052
Best to plot this on a piece of graph paper; "+i" is  to the right' , "+j" is upwards, "-i" is to the left, "-j" is downwards, all with respect to the origin.\n\n1. Draw the vectors on the graph paper. "u" is in the second quadrant (i is neg and j is pos). "v" is in the fourth quadrant, with (i is pos, j neg) . There are two answers to this question, since you could go from u to v or from u to v in the opposite direction. I always take the angle from zero to "u" and add it to the angle from 0 to "v".  For "u", the tangent of the angle to the vertical axis is 1/4, then angle = 14%. So the angle from zero degrees to the vert axis = 90, and "u"s angle from zero degrees is 90 + 14 = 104 degrees. For "v", the tangent of the angle to the horizontal axis is 4/3 = 1.333.. and the angle = 53 degrees. Therefore the angle from "u" to "v" is 104 + 53 = 157 degrees. (the other answer is 360 degrees - 157 degrees = 203 degrees if you rotate in the other direction).\n\n2. Again use graph paper. 240 degrees is 30 degrees less that 270 degrees (the vertical axis, down) draw a line 10 graph units long. It will be in the third quadrant where i and j will both be negative. At this point you can use a polar to rectangular converter, or do it by hand. By hand: the i side of the triangle formed with the down axis can be found by saying sin(angle) = i/10, and cos(angle) = j/10. The angle is 30 degrees (from vert downwards). so sin(30) = 0.5, and, cos(30) = 0.866; thus i= 0.5 and j = 0.866 (ignoring pos/neg). Since the vector is in the third quadrant, both i and j are negative.\nAnswer = -0.5i -0.866j\n\nHope this helps!
519965
Draw yourself a cartesian graph, vertical line is the y axis, horizontal is the x axis.\n\npick points on the x axis like 1, 2, 3, etc.  plug thos values into the equation and solve for y. then pair the numbers, place points in the graph by taking the x value going straight up (or down if y is a negative number) and place the dot accross from the y value.  do this for each set of points then connect the dots with a line or a curve. for instance in your equationstart with x=1\n\ny= 7(1-5)(1-2)^4(1+.5)(1+2.5)^3\ny= 7(-4)(-1)^4(1.5)(3.5)^3\ny=7(-4)(1)(1.5)(42.875)\ny=-1800.75\n\nso in XY format the first point would go at (1, -1800.75)
229964
According to Yahoo Dictionary, a normal (Mathematics) is A perpendicular, especially a perpendicular to a line tangent to a plane curve or to a plane tangent to a space curve.  So, bending towards a normal would mean it tends to go at a right angle from it source.
435860
First pair both x values are the same so it is a vertical line through -5 on the x-axis. Which would mean it has no slope (undefined)\n\nSecond pair has identical y-values so it is a horizontal line (slope=0) through 3 on the y-axis\n\nFor the third pair us the formula \ny=mx+c which describes any straight line (m=slope, c= a constant distinguishing the line from others with the same slope in different locations on a graph)\n3=m(-8)+c      (a)\n2=m(-6)+c      (b)\nSimultaneous equations!\nc=3+8m           (a)\n2=-6m+3+8m   (sub (a) into (b))\n2m=-1\nm=-1/2  \nSo the slope = -1/2\n\nThis formula can be used for any problem like this using the same method.
443970
y = f(x)\n\na) x --> x/3\nY gets a smaller angle, 1/3 to be exact.\n\nb) x --> -x\npoints downwards, instead of up.... (negative y)\n\nc) x --> 3x\nY gets a steeper angle (3 times)\n\nIf you have a graphical calculator, define 3 functions,\n1. f(x) = x/3\n2. f(x) = -x\n3. (x) = 3x\n\nThen you'll see what happens.\nYou can also use software for your computer to illustrate this, ex. winplot
433271
in the plane,\nthere are two kinds of curvature,\nthe regular one, which is the norm of the second derivative, if the curve has unit tangent.\nand the signed curvature, which in addition measures if the first derivative and the second derivative preserve orientation.\ns