How do you calculate a related rates problem when all three sides on a triangle are increasing?
958527
You need to know the angle between "x" and "y", call it "a"\nLet initial position of "x" be "x0"\nLet the time from start of "x" be "t"\nLet "z" be the distance between "x" and "y" at time "t"\n\nThen:  \ny = 38 + 2t\nx = x0 + 11t   \n\nAnd from Pythagoras:\nz^2 = x^2 + y^2 - 2xycos(a)\n\nHopefully that gets you started....
923704
Draw a diagram to represent the positions of the two ships at time t.  This will be a right triangle.  Side A will be the position of ship A and side B will be the position of ship B.  The distance between the ships will be the hypotenuse.  Let's call this S.  By the Pythagorean theorem,\n     S^2 = A^2 + B^2\nDifferentiate with respect to time, t.:\n\n2 S dS/dt = 2 A dA/dt + 2 B dB/dt\n Simplify by dividing the entire equation by 2\n\nS dS/dt = A dA/dt + B dB/dt\nSolve for dS/dt:\n\ndS/dt = 1/S (A dA/dt + B dB/dt)\n\ndA/dt is constant at 24 knots\ndB/dt is constant at 18 knots\n\nAt 5 PM,    A = 30 + 24(5) = 150\n                  B = 18(5) = 90\n                  S = SQRT( A^2 + B^2) = SQRT(150^2 + 90^2)\n                     = 174.9\nTherefore dS/dt = (1/174.9)(150(24) + 90(18)) = 29.8 knots.\n\nIt's just a coincidence that spoonish's answer is about the same as mine.  Only differential calculus can solve this type of problem.
980357
The answer has to be a speed!  It involves seeing that dy/dt = 1/2 x^(-1/2) dx/dt in this case.  Also, the distance from the origin is given by s^2 = x^2 + y^2; to get ds/dt (the required result), differentiate that wrt "t".  Put this all together yourself; it's the only way to learn rather than simply parrot.\n\nWithout a cast-iron guarantee, I get the value 27/(4 x sqrt 5) \ncm/s.\n\nLive long and prosper.
748474
Draw a triangle. Side a is 35m. Side b is 5m, and Angle C (the angle between sides a and b) is 155 degrees (180-25).\n\nsides a and b are the runners path. The remaining side, c, is the total displacement. By the law of cosines, the following is true of a triangle:\n\nc^2 = a^2 + b^2 - 2ab cos C.\n\nc^2 = 35^2 + 5^2 - 350 cos (155)\n\nAnd solving for c will get apprx  39.59 meters.\n\nYou can then plug this back into the law of cosines to get the angle B:\n\nb^2 = a^2 + c^2 -2ac cos B\n\n5^2 = 35 ^ 2 + 39.59^2 - 2 * 39.59 * 35 cos B\n\nSolving for B will get you about 3.06 degrees.
243812
At most shouldn't it take only 13 minutes?  If 1 & 2 cross together, then they will take 3 minutes, and if 3 & 4 cross together, they will need 10 minutes.  10 + 3 = 13.\n\nMaybe I don't really understand the question??
888670
First, draw a picture. Let C be Chris, B be Blaire, and A be the ball at some point. Thats a right angled triangle, where side BA is how far the ball has dropped; CB = 100 feet, and CA = the fishing line.\nNow, we need to know how fast the ball is dropping. Since its distance is 16t^2, its velocity is 32t (differentiating). So side BA is changing at 32t feet per second at any time.\nWe want to know the rate at which side CA is changing. But we know that CA^2 = CB^2 + AB^2, so CA = sqrt(CB^2 + AB^2) = sqrt(100^2 + AB^2).\nNow, we need to differentiate both sides with respect to time.\nThus, d/dt(CA) = d/dt(sqrt(100^2 + AB^2)).\nWe need the chain rule for the RHS: sqrt(something) differentiates to 0.5(something)^-0.5, then we multiply by the inside differentiated. AB^2 differentiated (with respect to t) is 2*AB *d/dt (AB) = 2*AB*32t.  So the overall equation becomes:\nd/dt(CA) = 0.5(100^2 + AB^2)^-0.5 * 2 * AB * 32t.\nNow, we know t = 3 and AB = 16*3^2, so substitute that in and you'll get the answer of 78.9 feet/second.\n\nI'm likely to have made a mistake, but hopefully that gives you the idea of what to do.
926663
the slope of the graph. the formula is rise/run. take 2 points X1,Y1 and X2,Y2  plug them in the formula X2-X1/Y2-Y1. youll get the rate of change! Best of Luck!
535960
The speed has no part in the solution, as it is the same for all three paths. So, you draw a triangly with legs 2.25 and 3 in lenght. the solution is the lenght of the hypotenusa formed. the answer is 3.75 min.
948476
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
766444
time to the minus one.\n\nDo a dimensional analaysis on a rate equation to confirm this.
929849
solve three body problems
1135961
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
763822
Divide the right angle into three equal parts, or three 30-degree angles.\n\nSounds like homework to me (which, hey, I'm cool with!), as if you owned this parcel this is NOT the way you'd want to divide it for real -- you wouldn't want three neighbors all bunched up like that at the confluence of the former right angle.\n\n.
663486
Diagram here:\n\nhttp://mathworld.wolfram.com/TriangularPrism.html\n\nto solve for its volume, calculate for the area of the triangle bottom and multiply it by the height of the prism.\n\nTo compute for surface area, compute for the area of the 3 rectangular faces and add the area of the 2 triangular faces.
93876
A quick guess without having reviewed your math, try adding g = 9.8 m/sec^2 to your answer and see if that is the answer that is needed.
1133022
It theoreticaly should not be too dificult to solve using vectors.  I am not going to solve it because I would prefer having exact values to work with, but basicaly, what you would need to do is find the complete distance, also make a graph where you find the time intersect of the torpedo and boat, and work with the x/y components to figure out what would be the ideal inclination.  It is a complex problem where you must work with more than one vector system, but it is doable if you work in parts.
780311
You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r