Find 2 points on the curve whose equation is given above, where the tangent line is vertical.
127996
(4,3) and (-4,-3)\nI think.\n\nCan't blame ya for asking about that one.  Trying to do that equation without a calculator before Calc III can be a pain since you don't get a shortcut in differentiating that type of  problem until late into Calc III.  It can be a pain even with a calculator without that shortcut due to the fact is doesn't balance easily either into a y=f(x) form.
327628
put it in standard form :\n\ny < -1/2x + 3/2 \nand \ny > 2/3x + 2\n\nthen you know the first line passes thru the y axis at 3/2 and the second at 2\n\nthe slopes are -1/2 and 2/3 .... then shade below the line on the first one cuz y is less than .\n\nthen shade above the second line because y is greater than
509021
a) y= 10^(tan (πθ))\n= e^(ln10*tan(πθ))\n= e^u where u = ln10*tan(πθ\ndy/dθ =dy/du * du/dθ\n= e^u * ln10*πsec²(πθ)\n= ln10*πsec²(πθ)*e^(ln10*tan(πθ))\n= ln10.πsec²(πθ)*10^(tan (πθ))\n\nb) y= √[tln(t^4)]\n= √[4t*ln(t)]\n= 2√[t*ln(t)]\n= 2u^½ where u = t*ln(t)\ndy/dt = dy/du * du/dt\n= u^-½ * (ln(t) + 1)\n= (ln(t) + 1)/√[tln(t)]\n\nc) y = arctan(arcsin (x^½))\n= arctan u where u = arcsin v where v = x^½\ndy/dx = dy /du * du/dv * dv/dx\n= 1/(1+ u²) * 1/√[1 - v²] * ½x^(-½)\n= 1/[2√(1 - x)*(1 + (arcsin (x^½))²)]\n\nd)xe^y = y - 1\ne^y + xe^y * dy/dx = dy/dx\nSo dy/dy - xe^y * dy/dx = e^y\nie dy/dx (1 - xe^y) = e^y\ndy/dx = e^y/ (1 - xe^y)\n= 1/(e^(-y) - x)
124576
first draw a right triangle and label the angles and sides. now angle B cut in half is 15 degrees and angle A is 90 degrees.  lets assign AM a value of x.  so the tangent function is equal to the opposite leg over ajacent leg=> tan(15degrees)=x/(2*sqrt(3)).  take the tan of 15 and then multiply that by 2*sqrt of 3 and that will be your answer for the length of AM which is approx=.9282  to find MC you must first find the length of AC with the same tangent function using 30degrees as your angle so lets assign AC a value of y. so tan(30degrees)=y/(2*sqrt(3)) which is = 2 so MC=AC-AM= 1.0718
365010
First you have to find the equation of the line that contains these two points (9,1) and (5,5)\nI believe the easiest way to do that is to use slope and y-intercept form (y=mx+b) m= slope and b= y-intercept\n\nto find slope, m = rise/run\nrise = -4\nrun = 4\nrise / run = -1\n\nTo find y-intercept plug in a coordinate pair and use are found slope.( i used 5,5)\ny=mx+b  x= 5 y=5 m=-1 b =?\n5=(-1)(5) +b\n10 = b\n\nSo this is our equation of the two points\ny=-x +10 \n\nNow we can use this equation to find whether  (10,0) is on  the line: Substitute (10,0) and see if the equation is still valid\n\n 0 = (-1)(10) + 10\n 0 = 0 Yes (10,0) is on the same line as the other points
390490
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
463997
Rewrite the line in slope-intercept form:\ny=2/3 x + 4/3\nSo you're looking for points on the graph of y=sec x such that dy/dx = 2/3\ndy/dx= sin x/cos² x = 2/3\nsin x/(1-sin² x) = 2/3\nsin x=2/3 - 2/3 sin² x\n2/3 sin² x + sin x=2/3\nsin² x + 3/2 sin x = 1\nsin² x + 3/2 sin x + 9/16 = 25/16\n(sin x + 3/4)² = 25/16\nsin x + 3/4 = ±5/4\nsin x = -3/4 ±5/4\nsin x = -2 or sin x = 1/2\n\nClearly, sin x can't equal -2, so you're looking for points where sin x = 1/2. There are two such points - π/6 and 5π/6 (i.e. 30° and 150° - you should remember them from your special right triangles). Therefore, your solution set is:\n\n{π/6, 5π/6}
28151
you find to points on the line \nexample (2,5) and (1,6)\naight m= slope\nput the points in this simple equation\ny2-y1 (over) \nx2-x1\n\n6-5 over\n1-2\nwhich Equals  \n M= 1/ -1\nHoped it helped\nor try asking the teacher for help after school or at lunch
516753
uh, find the derivative mathematically and then graph it.\n\nfor example to graph derivative of x^2.\n\nfind the derivative equation f(x)=2x\nand then graph y=2x.\n\nooops, didnt realize it was a picture.\n\nfind the zeroes (a high or low point i.e. tangent is a horizontal line)\nyou should be able to track poitive and negative slopes.  not too hard, you'll figure it out.
499735
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\n\n\nfor question 2:\ne^{sin(0.001)} is very similar to e^0.001 which in turn is approx 1\ns
296279
the parabola will be at the point 2,2 and shaped like a "u"
218797
the area under the curve f(x)!
556699
the slope or Tangent of the angle would be 9.75/17.24= or 0.5655. Now if you need to know the angle take tan inverse and that would be 29.5 degrees.
396034
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.
81300
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.
468139
All points on a vertical line have the same x-coordinate.  So, their equations are always x = that number.  Here you know that one point has an x-coordinate of -1, so all the others will too.  Therefore its equation is x = -1.
95542
There are two methods of solving this question.  One method is applying using the calculus, and other method is using Coordinate geometry.  Since u have not mentioned a particular method, easier way is applying the Calculus.\n\nLet (x1, y1) be a point on the curve.\nDifferentiating the given equation to find the slope of the tagent at (x1, y1), we get\ndy/dx = 6x -5\nAt (x1,y1) dy/dx = 6x1-5\n\nGive condition is the slope of the normal at this pt is = 2\nSo the slope of the tangent = -1/2\n\nTherefore, 6x1 - 5 = -1/2 which leads to x1 = 3/4\nSubstituting this value in the given equation we get,\ny1 = 3.(3/4)^2 - 5.(3/4)+1 \nThat is y1 = -17/16\n\nAs asked the abscissa is 3/4