Calculus Problem Help Plz!?
917983
A. Amount of Z in tank  = 4 lb\n\nB. After 10 minutes amount of Z = 4 lb\n\nC  Amount of water in tank at time t = (10 + 2t) gallons\n\nSo at time t, concentration of Z in tank = 4/(10 + 2t)\n= 2/(5 + t) lb/gallon\nSo 0.1 = 2/(5 + t) ie 5 + t = 20, t = 15 minutes
801052
It seems you are asking for f'(2)?  in which case, we have\nf'(x) = 1/(1+x^4)so that f'(2) = 1/9\n\nSteve
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
1102373
i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
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)).
945648
Take the derivative of f :\nf'(x) = 5x^4 (x+7)^4 + 4 (x+7)^3 x^5\nf'(x)= x^4 (x+7)^3 [ 5 (x+7) + 4x]\nf'(x)= x^4 (x+7)^3 [ 9x + 35]\n\nSo the derivative is equal to 0 for the values :\nx= -7\nx= - 35/ 9\nx = 0\n\nthe function is increasing in [-11;-7],\nthe function is increasing in [-35/9;+12],
912852
Let P be (a,b). The slope of the tangent at P is found from the derivative of the curve: -128 / x³, giving the equation of the tangent as\n\n          y - b = ( -128 / x³ ) ( x - a )\n\nNow find where this tangent lines cuts the X and Y axes. Let x=0 and solve for y. Let y=0 and solve for x to get\n\n          ( 0 , b + 128 / a² ) and ( a + a³b / 128 , 0 )\n\n(a, b) is on the curve so b = 64 / a². Use this to simplify the two intercepts to\n\n          ( 0 , 192 / a² ) and ( 3/2 a , 0 )\n\nUse Pythagoras to find the length between these points as the square root of\n\n          f(a) = 192² / a^4 + 9 / 4 a²\n\nNow the problem is to minimize this expression (ignoring the square root, because it is an increasing function). This is a standard minimization calculus problem. Set f '(a) = 0, and check f "(a) > 0 to find that the minimum occurs when a = 4√2, so the required point is \n\n          ( 4√2 , 2 ).
841586
(0,5/4)\n\nI solved this problem a few days ago using algebra.\n\nYou take the derivative of x^2, take the negative reciprocal (it will be -1/(2x)  This will be the equation of the line you are looking for.  The tangent points of the circle and parabola is perpendicular to the radius which goes through the center.  You have two different formulae for the slope, and the fact that the radius is one (use distance formula).  Then solve it and get the answer.\n\n\n\nSorry that's not by Maple, but it can be used to double-check your answer I guess.  If you want me to work it out in full, add a note and I will.\n\n\nHopes this helps.
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.
1032260
1. if x=f(t) and y=f(t) then\n\nx^4+y^4=6y\n\nd/dt { x^4+y^4=6y}\n\n4x^3 (dx/dt) + 4y^3(dy/dt) =6(dy/dt)\n\nand you have to have the x=f(t) & y=f(t) in order to substitute for the derivatives.\n\n2. if y=f(x) and x=f(t) then\n\nx^4+y^4=6y\n\nd/dt { x^4+y^4=6y}\n\n4x^3 (dx/dt) + 4y^3(dy/dx)(dx/dt) =6(dy/dt)(dx/dt)\n\nthen substitute for the derivatives of the functions y=f(x) & x=f(t)
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
657050
1) f'(x) = 40x^4+6x^2+4\nTo find local max's and min's, set this equal to zero, but you will see that all solutions will be imaginary, so there are no local max's and min's on the real line.\n\nf''(x) = 160x^3+12x\nTo find the points of inflect, set this equal to zer, but you will see that the only place where there is an inflection is a x = 0.\n\nSo what have we learned?  f(x) is a strictly increasing function, meaning tha there could only be one real root.\n\n2. Let u = 1+t and use substituion.
152408
Use excel to give delta y/delta x on the curve. Look for a point where this value (the derivative) shoots up in value. That's your point.
57176
http://www.calculus.org/
821550
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
819945
1.  the idea here is to take the limit for both pieces and show that they match.  This is relatively straight forward.  The limit as x approaches 1 from the left of x cubed is simply "1"     \nThe limit as x approaches 1 from the right of 3x-2 is 3(1)-2 also equals 1.  Since these two match, we have proven that "f is continuous at 1"\n\n2. The fact that f is continuous at 1 is INsufficient to imply that is is differentiable there\n\n3. the derivative from the right is simple 3 since the derivative of 3x-2 is 3.  The derivative from the left of 1 is also 3 since:\n\nf(x) = x^3 implies that f'(x) = 3x^2 and f'(1) = 3(1)^2 = 3 as well.  Since the derivative from the right equals teh one from teh left, the function is differntiable at x =1
190751
Hmm, well the limit of the secant line as delta-x approaches 0 is the tangent line, and 0.01 is a pretty small delta-x... so right now I would say the secant line is pretty close to f'(2).\n\nThat's all I can give you for now, but I'll think about it.\n\n~~~~~~~~~~~~~~~~~~~~~\n\nOk, well in that case, plug in 2 and 2.01 to get f(2) and f(2.01)\n\nf(2) = 2^2 + 5*2 = 4 + 10 = 14\nf(2.01) = (2.01)^2 + 5*(2.01) = 4.0401 + 10.05 = 14.0901\n\nNow you have 2 coordinates (2,14) and (2.01,14.0901) so to find the slope of a line between 2 points the formula is\n\n(y2 - y1)/(x2 - x1) --> note that y2 is the y from the 2nd coordinate, not y squared\n\n(14.0901 - 14)/(2.01 - 2) = \n\n.0901/.01 = 9.01