Question on differentials. Help pls!!?
813576
hypotenuse y=x/sin(t), x=9.5, t=26.75 degrees;\nAs dt=15’=0.25 degrees =0.004363 rad  is relatively small compared 26.75 degrees, then we’ll use 1st derivative: dy=-x*cos(t)*dt/(sint)^2, \n(a) p=|dy/y| = cos(t)*dt/sin(t) = cos(26.75)/sin(26.75)*0.004363 = 0.866%\n(b) dt=p*tg(t) or dt<0.02*0.5043=0.01086 rad = 0.578 degrees, q=dt/t<2.16%\n©  s=(1/2)*x^2*ctg(t), |ds|=0.5*x^2*dt/(sint)^2, r=|ds/s| = dt*tg(t)/(sint)^2 = 2*dt/sin(2t) = 1.0856% - check my calculations Rose, I’m poor with numbers!
299918
This is a separable differential equation\ndP/dt = 4P^2 + tP^2\nFactor P^2 out from the left side\ndP/dt = p^2(4+t)\nDivide each side by P^2 and multiply each side by dt\ndp/P^2 = (4+t)dt\nNow integrate each side\n-1/P = 4t + (1/2)t^2 + C\nSolve for P and you will get\nP = -1/(4t + (1/2)t^2 + C)
29068
pier_haringsma is correct.  You do not need any additional information.  Let \nD = total distance of the trip  \nD/2 = the distance the driver has gone so far\nT1 = amount of time he took to travel the first half of the trip\nT2 = the complete length of time of the whole trip (both halves).\n\n   Now, he wants the average ( D / T2 ) to equal 30.  His average so far is ( (D/2) / T1 ) = 15.\n\nD/T2 = 30, so D = 30*T2\n(D/2) / T1 = 15, so D = 30*T1\n\nD = 30*T1 = 30*T2, so T1 = T2.\n\n  So the driver wants the total time for the trip (BOTH halves), T2, to equal the time for the first half of the trip, T1.  This means he would have to travel the second half of the trip in no time, which is impossible.
795130
y = tan(πx - 1)\nlet u = πx - 1\nnow, dy/dx = (dy/du)(du/dx)\nd(tan(u)/du = sec^2(u)\ndu/dx = π, so \ndy/dx = πsec^2(u)\ndy/dx = πsec^2(πx - 1)
713982
use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
1017997
The equations are wrongor insufficient. Post the correct data if you want me to help/ Alternately, you could also message me.
1139627
dy/dx = 2x-3\nat X = 0. dy/dx = 2(0) -3 = -3
699429
c'mon dude, these are easy.\n\nAdd second to first.\n\n2x = 2, x = 1, therefore y = -2.
1001167
The solution to your question is a bit lengthy and it wont be easy typing that here. However I can tell you the guidelines to arrive at that.\n(1)Write down the equation\n(2) take differentiation wrt x on both sides.RHS will be 1 here.LHS will have to be diff..you must know how to diff e^(y/x).Use the diff rule for products.\n(3) take diff.again of the above obtained equation.you will get alot of terms then.try getting from that the form you want.
888459
v(t)= -gt - ve(ln((m - rt)/m))\n\nx(t) = ∫v(t) dt\n\n= ∫[-gt - ve(ln((m - rt)/m))] dt (Note ∫lnu du = ulnu - u + k)\n\nLet u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du\n\nSo x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c\n= mve/r [ (m - rt)/m * (ln((m - rt)/m)  - 1] - ½gt²\n= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²\n\nIf g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s\n\nx(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60²\n\n≈ 138390 m\n\n= 138.39km
400622
This is a great question to illustrate the calculus principle of differentiation (i.e. differential of y with respect to x)\n\nx2-x+m=0 ---- eqn. 1\nx3-x2+6x+24=0 ---- eqn. 2\n\nSince both are = 0, for eqn. 1 = eqn. 2, we have:\n=> x2-x+m = x3-x2+6x+24\n=> x3-2x2+7x+24-m=0\n=> x3=2x2-7x+m-24 ---- eqn. 3\n\nLet y = x3. Therefore, in eqn. 3, we have:\n=> y = 2x2-7x+m-24  ---- eqn. 4\n\nTaking the differential of y w.r.t. x (i.e. dy/dx), we have RHS (right hand side):\n=> dy/dx = 4x-7 (since m and 24 are integers, they 'disappear' as they do not vary with x) ---- eqn. 5\n\nBut from eqn. 4, we know y = x3. Therefore, LHS(left hand side):\n=> dy/dx = 3x2 ---- eqn. 6\n\nSince eqn. 5 = RHS and eqn. 6 = LHS, we have:\n3x2 = 4x-7\n\n3x2 -4x +7 = 0  (in the form of ax2+bx+c=0)\n\nTechnically, there is no real solution for x (perhaps you have input a wrong sign for one of the 2 equations in your question?) Anyhow, proceeding, using general formula for quadratic equations where x = 1/2 (-b +/- sq.rt of [b2-4ac]), then:\n=> x = 1/2( 4 +/- sq. rt of (-80))\n     = 2 +/- sq. rt. of (-20)\n    = 2 +/- sq.rt. of 20 i (where i = sq. rt. of -1)\n\nAbove topic is calculus topic on REAL AND IMAGINARY numbers.\n\nTherefore, subbing values for x into eqn. 1, you will get the 2 values for m.\n\n\nOR alternatively:\nBy trial and error on eqn. 2, we get:\n=> x3-x2+6x+24 = (x+2) (x2-3x+12)\n          \nSolve values for x = -2, 1.5 +/- sq.rt. of 39 i (where i = sq. rt. of -1)\n\nSubstitute values for x into eqn. 1 and you will get values for m.
981618
I guess 1) just wants you to write: Want dA/dt, have da/dt and db/dt.\n(From now on I'll use A' to mean dA/dt etc).\n2)\nA=π*a*b, so using the product rule, A' = πa'b + πab'. So A' = 7πb - 11πa.\n3) Just substitute in: A' = 7*π*50 - 11*π*30 = 20π, so thats the rate it is increasing.\n4) Again substitute in, A' = 7*π*20 - 11*π*40 = -300π, so now its decreasing at that rate.
765615
differentiate implicitly:\n\n2x + 2yy' = 0\n\nand then some algebra\n\ny' = -(x/y)\n\nthen differentiate again:\n\ny'' = (-y - (-x)*y')/y^2 = (-y + xy')/y^2\n\nthen substitute in values:\n\ny' = -0/5 = 0\n\nand y'' = (-5 + 0*0)/(5^2) = =-5/25 = -1/5
160821
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
806346
using implicit differentiation:\nyou get 2y(dy/dx) - [y + x(dy/dx)] -3 = 0\nthen move 3 over and factor out (dy/dx):\n(2y-x)(dy/dx)=3 + y\ndy/dx=(3+y) / (2y-x)\nnow plug in your point (0,-1)\ndy/dx = (3+ -1)/(2(-1) - 0\ndy/dx = -1\nanswer = -1\n\n(dont forget to use your chain rule when you are implicitlly differentiating)\n\n#2\nsame thing make sure you sue multiplication rule\n[y^2 + 2xy(dx/dy)] - 2(dx/dy) + 12y^2(dx/dy) = 0\nthen move over the y^2 and factor out dx/dy\n(dx/dy)[2xy - 2 + 12y^2] = y^2\ndx/dy = y^2/ (2xy -2 +12y^2)\nPlug in 1 for y\ndx/dy = 1/(2x -2 + 12)\ndx/dy = 1/ (2x +10) or (2x + 10)^(-1)\n\nAnswer (leave in terms of x) = 1/ (2x +10) or (2x + 10)^(-1)\n\nI may have made mistakes so let me know
833002
Let u = x², so du = 2xdx\n\nSo, ∫ (x/1+x^4) dx =\n\n1/2 ∫ (1/1+u²) du =\n\n1/2 arctan(u) + C = \n\n1/2 arctan(x²) + C\n\n~ ♥ ~