How do I find the equation of the tangent line...?
918737
The equation for the tangent is:\ny = f'(a) * (x-a) + f(a)\n\nYou need to calculate the derivative h'(t) for t=3s and plug that into the equation, adapted to your problem:\n\ny = h'(3) * (t-3) + h(3)\n\nI'll let you do the maths...
274298
Use implicit differentiation to find dy/dx. Standard differentiation rules apply, except when you differentiate a term with a "y" in it you must tack on a dy/dx, so:\n\n2x+8y(dy/dx)=0\ndy/dx = -x/4y\n\nThe tangent line intersects the ellipse at (x1,y1), so the slope of the tangent line will be -x1/(4*y1).\n\nWe can also compute the slope of the tangent line by plugging the two points (-5,0) and (x1,y1) into the standard slope formula m = delta(y)/delta(x) and get that the slope of the tangent line is y1/(x1+5). Since these expressions both describe the slope of the tangent line, they must be equal, so\n-x1/(4*y1)=y1/(x1+5)\nx1^2+5*x1=-4*y1^2\nx1^2+4*y1^2+5*x1=0\n\nSince (x1,y1) is a point on the ellipse, it must satisfy the equation of the ellipse, so x1^2+4*y1^2=5. Substituting this in to x1^2+4*y1^2+5*x1=0 gives us 5+5*x1=0, and x1=-1. At this point, there are actually two y1 values that will satisfy the equation of the ellipse, y1 could be either plus or minus 1 (get these by letting x=-1 in the original equation of the ellipse), and so there are actually two points on the ellipse where the tangent line passes through (-5,0), and those points are (-1,1) and (-1,-1).
358399
There is an equation for this, but I don't remember what it is.  Guess and check with a calculator is funner anyways...\nI didn't show much work (how much work is there with guess and check), but I think these are correct. (&=infinity)\n\n1. The range of k is [-2,&)\n\n2. When k=3 then the ratio of a:b will be 1:5.  The equation of the parabola will be y=-X^2 + 4X + 5.\n\n3. I wasn't sure exactly what you were asking on Q 3.  If you are looking for where the tangent line of the parabola at x=0 crosses the x-axis, the point will be (-1.25,0).  If you are saying that from (A,0) to (B,0) in Q 2 is the diameter of a circle, and you are looking for where the tangent line of that circle at x=0 intersects the x-axis then the point is (-4.5,0).\n\nI hope that helps.
856434
First, look at where the curve intersects the line:\n\ncurve y=1/x\nline y=x/4\n\n1/x=x/4\nx=+/- 2\n\nNote that the curve is symmetrical about the origin for x&y positive and x&y negative, so the line intersects twice.\n\nThat's actually somewhat irrelevant, but it helped me visualize the line and the curve.\n\nTake the first derivative of the curve\ndy/dx=-1/x^2\nThis is the instantaneous slope of the curve at any point, which is the tangent by formal definition.\n\nA line perpendicular to the line y=.25x will have a slope of -4.\nSo set the derivative equal to -4 and solve for x\n-4=-1/x^2\nx=+/- 1/2\nsolve for y\nand find the points\n(1/2,2)\n(-1/2,-2)\nThis is where the two lines that are both tangent to the curve and perpendicular to the line intersect the line.\n\none line is y=-4x+4\nthe other is y=-4x -4\n\nj
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
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).
786513
put x^2=t\n2xdx=dt\nxdx=1/2dt\nthe integral now is\n(1/2dt/1+t^2\n(1/2)tan^-1t+C\n=(1/2)tan^-1(x^2)+C
772313
You draw it under the curve\n\nThe tangent line will only touch the curve at one point for this type of smooth upward bending curve.(At least for the part of the graph shown).\n\nYour graph plots total displacement verses time. If the graph was a straight line then this would indicate a constant "velocity" but since the curve bends upwards you are dealing with a positive acceleration.\n\nPractically, from your graph, suppose you wanted to know the approximate slope of the tangent line at t=5. Use the two closest data points: t=4, d=8.4 and t=6, d=16.5. Then the slope is about (16.5-8.4)/(6-4) = 4.05 at t=5. it will be slightly greater than this. You can use a straight edge to make a visual adjustment and then you can find the slope by dividing the displacement value at t=5 by (5 minus the value of t where the straight edge crosses the time axis).\n\nThe other way is to use the data to deduce, a closely as possible, the formula of the function being graphed. For example if the displacement was directly proportional to the square of the time, d = k t^2 for a determined value of k, then the slope of the tangent line at any time, t, would simply be 2 k t which is the time derivatiive of the function.
909269
Take the derivative of the original function. For your given example, if I remember my calculus correctly:\nf(x)=y\nf'(x)=1\nGraph the new equation.
184556
Let x(t) = A*cos( w*t + theta )\n\nthen x(0) = A*cos( theta )\nand dx/dt (0) = -A*w*sin(theta)\n\nso if you know x and its derivative in t=0 u have two equations and two unknowns ( A and theta ) . the rest is simple algebra .
831486
Differentiate both sides, we have \n\ndy =  (1/x)dx + (1 + tan^2(y))/tany dy\n\ntherefore\n(1 - (1 + tan^2(y))/tany)dy=(1/x)dx,\n\ni.e.\n\ndy/dx=tany/(x(tany - tan^2- 1)).
768749
A'ighty, first write sec(x) = 1/cos(x).  Now take the derivative using the Quotient Rule:\n\nd/dx (1/cos(x)) = (cos(x)*d/dx(1) - 1*d/dx(cos(x))) / cos^2(x)\n\n                        = (0 - (-sin(x))) / cos^2(x)\n\n                        = sin(x) / cos^2(x)\n\n                        = (1/cos(x)) * (sin(x) / cos(x))\n\n                        = sec(x)*tan(x).\n\nAnd there you go.  ^_^
748469
The first thing I would do is rewrite both a and s in terms of h.\n\nThen you use the following rules to find the derivative.\n\n1.  Use the quotient rule to find the derivative of  sin(a)/s^2.  If you don't know the quotient rule -- look it up and memorize it.\n\n2.  The derivative of the sin is the cosin -- use the chain rule to get cos(a)*da/dh\n\n3.  The derivative of the bottom is  2*s * ds/dh (again using the chain rule).\n\nOnce you get the derivative, solve for zero.\n\nGood luck.
730760
θ = α + β  where α = tan^-1 (1/3) and β = tan^-1 (1/2)\nie tanα = 1/3 and tanβ = 1/2\n\ntanθ = tan(α + β)\n= (tanα + tanβ)/(1 - tanαtanβ)\n= (1/3 + 1/2)/(1 - 1/3 * 1/2)\n= (5/6)/(5/6)\n= 1\nSo θ = ....., π/4, 5π/4, .....\nSo cosθ = ±1/√2 = ±√2/2
291468
Oh yes i have cracked it at last....\n\nsolution...\n\nLet x=tan-1 (y) (i mean tan inverse)\nthen tan x = y,and cot x = 1/y (coz tan-1 (x) = cot-1 (1/x))\n\nthen ur given equation's Left hand side part (LHS) will be\n\n(1/y)/(1-y) + (y)/(1-(1/y))\n\n= 1/(y-y2)+ y/(1-(1/y))\n\n= 1/(y-y2) + y2/(y-1)\n\n= y2(y-1) - 1/y(y-1)\n\n= y3/y(y-1)\n\nI hope u understood till this step.\n\nNow we have (a3-b3) = (a-b) (a2+ab+b2)\n\nthen the LHS becomes...\n\n(y-1)(y2+y+1)/y(y-1)\n\n= (y2+y+1)/y\n\n= y+1+(1/y)\n\nsubstitute back y=tanx, 1/y = cot x and leave 1 as it is...\n\nU have LHS = 1 + tan x + cot x...Thats ur RHS !!!
805503
need to take the derivative vector:\n( -3 sin t, 4 cost)\nwhen t= pi/4, the vector is: \n(- 3 sin(pi/4), 4 cos(pi/4)) = ( -3/sqrt(2), 4/sqrt(2) )\nnow, the slope of the tangent line is:\n (4/sqrt(2) ) /  ( -3/sqrt(2) ) =-4/3\nso the equation of the tangent line is:\ny- 4sin(pi/4) = -4/3 (x-3cos(pi/4)) \ny-4/sqrt(2) = -4/3( x-3/sqrt(2) )\n'
224079
try this site... it shows step by step how to go about sloving these equations, good luck: