Parabolic Bridge Math Question?
412025
3.36 meters.  \n\nHere's why:  find the formula for the parabola.  We know that when X is 0, Y is 4.  And, we know that when X is 2.5, Y is 0.  So the formula is:\n\nY = 4 - (4/6.25)*X^2.\n\nNow let X = 1 (half the width of the truck), you'll see Y is 3.36.
923403
Start off by drawing the picture. You have a right triangle, with a hypotenuse of 25, a base (b), and height (h). \n\nDon't get jumbled down by numbers yet, ignore them. Visualize what you are calculating.\n\nVelocity / speed is the derivitive of position. (v = delta x / delta t). So all you need to do is find an eqation relating the geometry using known and unknown quantities and then differentiate with respect to time. Sounds confusing but all you have to do is....\n\nYou know, a^2+b^2 = c^2 for a right triangle. In our problem,\n\nb^2+h^2 = 25^2\n\nThis is our position equation. Now differentiate with respect to time.  DO NOT forget the CHAIN RULE! <------ VERY important\n\n2*b*db/dt +2*h*dh/dt = 0.\n\nFrom here just use algebra; where,\n\nb=7 ft, and db/dt = 2 ft/sc.\n\nWhich leaves two equations and two unknowns.\n\nThe important thing to remember is to take this step by step and try to visualize what you are calculating. Also, looking at the units of the given quatitities can give you some idea of what you will be required to do. \n\nAlso, just remeber that speed / velocity is the derivative of position with respect to time (the units reflect this).
782406
°oO****Oo°°oO****Oo°°oO****Oo°\n\n\n 4 ∫[g(x)] g'(x) dx = 4 x 1/2g(x)² +c = 2g(x)²+c\n\nits the first suggestion that u made but i think you forgot the number 4...\n\n\n°oO~<<_P~'*'_ I~'*'_M~'*'_O~'*'~L _~'*'_A>>~Oo°
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
1053965
Simple parabolic path formula \n\na hit ball is given a velocity v at and angle theta.\n\nAssuming no air resistance the balla travels through the air along its horizontal vector un abstructed, while acted on by gravity in the vertical direction.\n\nthe equation\n\nd=vt + 1/2gt^2 can be solved for d which will be the total vertical distance at time t, where t is the time at which the ball is the highest, lets say about 1/2 of the total time, g is the gravitational constant and v in the velocity in the vertical direction
759773
Let x be the larger number\nLet x - 100 be the smaller number\nThe difference between the two is 100.\n\nThe product would be x(x-100) = x² - 100x.\n\nIf you take the derivative, you get 2x - 100.\n\nSet this to zero:\n2x - 100 = 0\n2x = 100\nx = 50\n\nAlternatively, if you aren't familiar with calculus, turn the equation into vertex form for a parabola.  Do this by completing the square:\nx² - 100x = 0\n\nTake the coefficient on the x-term (-100), divide it by 2 (-50), and square it (2500).  Add this to both sides:\nx² - 100x + 2500 = 2500\n(x - 50)² = 2500\n(x - 50)² - 2500 = 0\n\nThis is vertex form a(x - h)² - k = 0 and the vertex will be at the point x = h or x = 50.\n\nSo either way you figure it, the numbers are 50 and -50.  The product is -2500 which is minimum.
929836
v = b - u/k\nv - b = u/k\nk(v-b) = u
901005
Use the following equation and solve:\n\nP1V1/T1=P2V2/T2\n\nT2=P2V2T1/P1V1\n\nalso note the significant figures!!!!
902316
Let y be required distance from the far end.\n\nThe weight is acting at the C.G i.e., at 2 m from the far end.\n\n2/3 of the weight w is acting upward at y and \n\n1/3 of the weight w is acting upward at 4m from the far end.\n\nTaking moments about the far end, \n\n2/3 W y + 1/3 W 4 = W x 2. \n\n Dividing both sides by 2W and multiplying by 3 on both sides\n\ny + 2 = 3 \n\ny = 1m.
723707
s'(t) =(t-3)(-2t^{-3}) + t^{-2}\n=( -2t+6 +t)/ t^{-3}\n=(6-t) / t^{-3}\nso the velocity is 0 when the top =0\ni.e 6-t-0\nt=6\ns
1097253
i'm not an engineer but i would say that there are more than one methods. i think you have to make each carrier autonomic,which means it can react as an indepenent part of the bridge in forces like wind or earthquake. i don't know if i helped, i hope i'm not beyond the subject of your question? if i'm sorry
590082
1) Let's think about this problem two-dimensionally.  A cross-section of the parabolic mirror would simply be a parabola.  Let's translate this onto a Cartesian coordinate system with (0,0) as the center.  Then (0,35) would be the focus, y=-35 would be the directrix, and one of the upper points would be (27,y).  If we find y, we find the depth of the mirror.  Since the points on a parabola are those points that are equidistant from the focus and directrix, let's set these distances equal to one another.  The distance between the focus--(0,35)--and the top point--(27,y)--is given by sqrt((35-y)^2 + 27^2).  The (shortest) distance between the directrix--y=-35--and the top point--(27,y)--is simply y+35.  Setting these two distance equal, we solve for y:\n\nsqrt((35-y)^2 + 27^2) = y+35\n(35-y)^2 +729 = (y=35)^2     (square both sides)\n1225 - 70y + y^2 + 729 = y^2 + 70y + 1225       (simplify)\n729 = 140y    (simplify)\ny = 729/140\n\nThus, the depth of the mirror is 729/140 ft.\n\n\n\n2) We must assume this 6% is an annual interest rate.  Then, the problem is simple.  A = P * (1 + r/n)^(nt), where A is the ending amount, P is the principal, r is the annual percentage rate as a decimal or fraction, n is the number of times compounded per year, and t is the number of years.\n\nIn this example, P = 6000, r = 0.06, n = 4, and t = 10.  Plugging these numbers into the above formula, we find that A = 10884.11.\n\nThus, after ten years, there is $10884.11 in the account.\n\nGood luck!
865937
w=x\nl=2x\n2w+2l=P\n2x+2(2x)=2\n2x+4x=2\n6x=2\nx= 1/3\n\nwidth is 1/3 and length is 2/3
1107625
I assume you are able to take partial derivatives. So take the partial derivatives that appear in the heat equation (you did look up the heat equation, right?) and plug them in to see if you get equality.\n\nSo, the partial derivative with respect to t is\n-lambda^2 k*u(x,t). This is the same as k times the second partial derivative with respect to x. Looks right to me!
200344
W=f.delta(x) \nx=w/f = 49821/717= 69.49 metes
613149
Firtly 1 And 2  cross the bridge and 1 takes flashlight back.Time Spent=2+1=3 then 3 and 4 cross the bridge amd 2 takes flashlight back.Time Spent 10+2=12.And now 1 and 2 can cross the bridge.Time Spent=2.Total time spent=3+12+2=17.
144053
Pay more attention in class! And assuming it is an ideal gas, use PV=nRT. Since there is not change in volume you can basically drop it from the equation and make P=nRT. If you like algebra you can leave it in but it will cancel out eventually. You could solve for 'n' first using initial P and T, and then plug and chug.