The first and second derivative test and curve sketching?
823012
Either one is possible. Consider y=x² (one critical value, which is not an inflection point) and y=x³ (one critical value, which IS an inflection point).
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
826809
It is probably best to determine the range after you have the graph.  You will need to look at the local maximums, local minimums, and the horizontal asymptotes to determine the range.  (In other words, it is not a fast problem).
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.
907697
You need to find an equation first. try drawing the box out it may help. Draw a rectangle label one side 16 inches the other 30 then draw squares at each corner.\n\nLet x be the length of one of the square's sides\n\nVolume = length * width * height\n\n= (30-2x) * (16- 2x) * x\n\n=4x^(3) - 92x^(2) +480x \n\nTherfore  for largest volume: maximise V= 4x^(3) - 92x^(2) +480x \n\nTo maximise: first find values of x when the first derivative is equal 0:\n\nV ' = 12x^(2) -184x + 480 = 0\n\nx = 12 or 3 1/3\n\nNext verify which gives a maximum V value by either using first derivative test or second derivative test:\n\nWith first derivative test we choose values between intervals of x values for V'=0 and look at the trend:\n\nPositive values mean the function is increasing (/) and negative values means the function is deacreasing at that point (&#92;)\n\nx | 0 | 3 1/3 | 5 | 12 | 14                  \nV'| / .|... - .. | &#92;. |.. - ..| /\n\nTrend on either side of x= 3 1/3 :/&#92;\nTrend on either side of  x=12 : &#92;/\n\nTherefore max is at x= 3 1/3\n\nBy second derivative test :\nIf v"< 0 ie a negative number, V is a maximum\nIf v">0 ie positive number,V is a min\nIf V" = 0 it is a point of inflection\n\nV"= 24x-184\n\nx= 3 1/3,  V"= negative therefore max\nx= 12 ,    V"= positive therefore min\n\nHence the squares should be of lengths 3 1/3 inches to give the largest volume
160190
the second part is graph thy.
826981
It looks to me like it's .5d which is not too far from your answer.\n\nFor simplicity, use a unit of length in which d = 1 and a unit of time in which the speed of each of the two bugs is also 1. Let r be the distance between them and let h be the angle between the y direction and the vector from S to R. In other words h starts out pi/2 and approaches 0.\n\nThe time derivatives are:\n\ndh/dt = -sin(h)/r , dr/dt = -1 + cos(h)\n--->\ndr/dh = r(1-cos(h))/sin(h)\n\nSeparating the variables r and h to opposite sides of the equation and integrating both sides with the initial condition (r=1 at h=PI/2) gives:\n\nr = tan(h/2)/sin(h)\n\nYou want the distance between the two bugs as t-->oo or equivalently as h --> 0 which is r=1/2 (use L'hopital's rule for example), or d/2 in terms of the initial distance d.\n\nOf course S never actually reaches the y axis but approaches it asymptotically.\n\n(Maybe I should add, I first tried to formulate the problem in terms of the Cartesian coordinates and was able to write a differential equation for the curve but it looked very difficult to solve. Finally tried using the above coordinates (r,h) and it became much easier)
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
1177010
If the you have a local max/min at an interior point, , the partial derivatives must be zero.\nSince the sum of the second partials is zero, the point must be a saddle point, not a max/min.
787521
Example:\nIf z=f(u,v),\nu=g(x,y) and\nv=h(x,y), then\n(all derivatives are supposed to be partial derivatives)\ndz/dx=dz/du*du/dx+dz/dv*dv/dx\ndz/dy=dz/du*du/dy+dz/dv*dv/dy\n\nAlternatively, you can set of the matrices\n[[dz/du dz/dv]]\nand\n[[ du/dx du/dy]\n[dv/dx dv/dy]]\nand get that\n[[dz/dx dz/dy]]\nis the product of the two matrices.
636329
I like to use GraphCalc, as it also includes live rotation of 3-D graphs. See http://www.graphcalc.com/ .
158569
Here is a tutorial...you can change any of the parameters that you mentioned and look at the graph as it is made and changed...really cool if you ask me.\nhttp://www.analyzemath.com/logfunction/logfunction.html
466006
Pick a few easy points like pi/2, pi, 3pi/2, 2pi.  You know the period.  Then just graph it.
784392
Here is what i found.  I'm not sure how right this is, but you can check it out.  Good Luck with your testing.
723693
I'm pretty sure you are correct.  I am also working thru Calc III (at the moment, actually).  It helps to picture the gradient as perpendicular to the tangent plane.
799961
google's sketchup.  It's free and easy to use!
697299
Graph them both.  Notice that, after graphing the one, you only need to turn your "page" 90 deg counterclockwise and you'll get a mirror image of f^-1(x).  \n\nYou'll also notice that each has exactly one rectangular area that the other does not.  The first, f(x), is a "piano"-shape, formed by a "pseudo triangle" and a rectangle, the latter of which measures 8x5, because 8=12-4.  Now take that rectangle away, and look from the point of view of f^-1(x).  This "triangle" lacks a rectangle 4x4 in dimensions, where the second 4 is the height, from f^-1 (9) = 4, and the first 4 is from 4=9-5.\n\nThus, we subtract 8x5=40, and add 4x4=16, to get our answer (which I'm actually too lazy to finish giving; it's 24 less than the first area).