Where the tangent is perpendicular to...?
922182
The slope of y=1-4x is -4, so the slope of its perpendicular is 1/4. Thus, we must find a point where dy/dx = 1/4. Taking the derivative:\n\ndy/dx = 1/x²\nSetting this equal to 1/4:\n1/x²=1/4\nx²=4\nx=±2
380640
F.G is the component of F laying on G.\nl=F.G/g=(10*0.1-6*0.2+5*0.3)/g\nl=(1-1.2+1.5)/sqrt(0.1^2+0.2^2+0.3^2)\nl=(1.3)/sqrt(0.14)=3.5\nNow the component of F perpendicular to G is:\np=sqrt(f^2-l^2)=sqrt(100+36+25-12.25)=13.2\n\nThe answer is 13.2\n\nthe direction is parallel to: (G*F)*G\nwhere * is cross product.
893970
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`
1186712
arctan x should be interpreted as "the angle whose tangent is x". So draw a right triangle and include an angle whose tangent is x. Then figure out the sine of that angle in terms of x and you're done.\n\nEasiest way to do this is to let the side opposite the angle be x and the side adjacent to the angle be 1. Then the hypotenuse will have length sqrt(1+x^2).
488283
I think that what you really want here is the tangent ratio. You know the slope of the line, right (hint, it is right there in the equation)? You know that the tangent ratio is a over b (where a and b are the non-hypotenuse sides of the triangle), right? You should be able to use that to figure out the tangent of the angle made by the line. Now all you have to do is figure out the angle that has that number as a tangent. That would be the "tan -1" (inverse tangent) function on your calculator.\n\nIf you are working from trig tables rather than a calculator, find the number in the second column of the tangent table that is closest to your tangent number then look in the first column for the angle.\n\nIf you are required to use sin and cosine remember your Pythagorean theorem and use it to find the hypotenuse.\n\nASIDE TO Drakestone: Your solution only finds the sine of the angle, not the angle itself. You need to apply "sin -1" to your answer to get the final answer.\n\n\nASIDE TO eternal...: You are wrong too. graph y=5x+3 and you will see that it is way too steep to be an 11 degree angle. You mixed up your opposite and adjacent values in calculating your tangent.
897760
The slope of the tangent is 2x, so the slope of the normal line is -1/(2x), not -1/2 x, which you used later.\nIf you continue of, you get -1/(2a), so the equation is y = -1/(2a) * x + b.\nSo a^2 = -1/(2a) * a + b = -1/2 + b.\nThus b = a^2 + 1/2.\n\nIn this case, theres no dividing by a, so you can just substitute in a=0 if you wanted (that was a good hint that your answer was wrong).
534053
It really depends upon your problem.\n\nIf you are asked to find the sine, cosine, and tangent then for any right triangle:\n\nsine = opposite side/hypotenuse  = y/r\n\ncosine = adjacent side/hypotenuse = x/r\n\ntangent = opposite side/adjacent side = y/x\n\nwhere opposite side is the side opposite the angle in consideration.\n\nwhere adjacent is the side closest to the angle.\n\nwhere hypotenuse is the longest segment in the triangle.
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 ).
691760
y = x\nFirst do implicit differentiation to solve for y'. Next, plug in 1 for x and y to find the numerical value for y' which is the slope for the TANGENT line. The normal line will have a slope that is the negative inverse of the tangent line's slope. The tangent line's slope is -1 so the negative inverse is 1. Now that we have the slope of 1 for the normal line, one can use the formula y = mx+ b to find the intercept. The intercept in this case happens to be zero so the final answer is y = x.
105143
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma
197729
Are you sure you wrote the problem right?  It is at y=3 and not x=3?\n\nA previous poster solved for y(x) when x=3 and got 38 (which is correct) but that's not what you asked.\n\nI think you will have to solve x^3 + x^2 = 3 to get a value for x before you can go on.\n\nAfter that, it's pretty straight forward.  You would find y'(x) to get your slope.  Then the equation of the tangent line of f(x) at x=a is y=f(a)+m(x-a).
1075301
find the slope, m, of your line (-2/3 in this case). \n\nThe slope of any perpendicular line will be -1/m (3/2 in this case).\n\nSo for any real number b,\n. . . . . . . y = (-1/m)x + b = (3/2)x + b\nis the equation of a line perpendicular to the original line.
826563
multiply the tangent of the given angle by the known side as long as it's not the sloping side (i.e. the hypotenuse). You will need a scientific calculator to do this. I am assuming of course that you are dealing with a right triangle. This will give you the other side of the triangle. Example:\n\nlets say you have side A of a right triangle and an angle of 30 degrees and A = 10. Take your scientific calculator press 30 and then the tan button depending on the type of caculator you have. Now multiply that result by 10 and you will get side B.\n\nSide C is found by A^2 + B^2 = C^2. take the square root for your answer.\n\nNow if the given angle is opposite the given side. Just calculate the angle of the given side by adding the given angle to 90 degrees and subtract the result from 180, and then multiply the tangent of the new new angle by the given side.\n\nTheoretically you can check the answer by multiplying the new side by the tangent of the new angle. You should in theory get the given side. But my trig is so rusty I may have forgotten more than  you will ever know. good luck.
1171148
the portion  between the two vector that happened to be perpendicular is BSin(alpha)\nalpha is the angle between the two Vectors, Alpha  is   \n ArcTan(B/A)\nWhere B & A are the two vectors.
846631
Construct meaning with a compass and straightedge? And by "adjacent" they're already at right angles? Just measure each side with the compass, move it so one leg is at the other corner, and draw a small arc approximately where the other leg extends. Do this for both sides, then the point of intersection is your missing point, and you can draw out the rectangle with the straightedge.\n\nIf the sides are not at 90 degrees, just turn one of them (measure one with the compass, draw a circle based on it, then take the two points of intersection and construct the perpendicular bisector, then take the point where that intersects....)
916414
For a line to be perpendicular all you need is the negative inverse slope.\ny=2x-3 has a slope of 2 so the negative inverse would be (-1/2)\n\nSo y=(-1/2)x-3 would cross perpendicular at the y axis at -3
764932
slope = f'(0),\nf'(x)= e^x 1/(x^2 +3) 2x + e^x ln( x^2 +e)\nf'(0)= e^0 1/3 (2(0)) + e^0 ln(e)\n= 1\n\nnow y=f(0)=e^0 ln( e) =1\nso the eq. of the tangent line is:\n\ny-1= x\ny=x+1\ns