Calculus question involving a cylinder.?
817821
It is my understanding that you would multiply the given costs by the surface area of those pieces to get total cost (C). Your equation would be as follows:\n2[(πr^2)(.027)] x πrh(.015)=C\nThe constraint equation is the volume:\n(πhr^2) = 16\nsolve your constraint equation for one of the variables.\nThen take the derivative of the cost equation and solve for your variable at "0." This will give you r (where r is radius) or h (where h is height of can) depending on what you solved for.
427111
This is essentially a washer. The outer radius, R, is the diffference between y=6 and the x axis (y=0) or 6. The inner radius r, is the difference between y=6 and the equation y=4x-x^2. The x axis and the function intersect at x=0 and x=4\n\nThe volume ,V is given by the integral\n\nV= pi int. (from 0 to 4) (R^2-r^2)dx\n \nor v= pi int. (from 0 to 4) ((6)^2-(6-(4x-x^2))^2)dx
750653
The rotation rate is such that the centripetal acceleration equals the component of the downward acceleration of gravity that acts normal to the cylinder's wall at 68 deg. Thus, solve the following for w (rotation rate):\nw^2*r = g*sin(68).\nSince you asked what equation to use and did not ask for a worked-out answer, that's it. I hope it helps.
549879
haha I was just working on my geometry homework and we're doing that right now.. the formula for the volume of a prism is: V=Bh, where B=the area of the base and h=the height of the prism.. that's the equation for ALL prisms..\n\nthe equation for the volume of a cylinder is V=pi(r^2)h\nso it's pi times the radius squared times the height\n\nthe equation for the circumference of a circle is C=2(pi)r\nso 2 times pi times the radius
863660
Use volume formula for a cylinder:\n\nV = pi r² h\n\nYou know r = d/2 = 18/2 = 9 ft.\n\nYou know h = 47 ft.\n\nSo, using 3.14 for pi as your question requests, you get the following:\n\nV = 3.14 * (9 ft.) ² * 47 ft.\n\n= 3.14 * 9 * 9 * 47 ft ³\n\n= 11,954 ft ³ (rounded)\n\nHope that helped!\n\n.
401455
Sorry.  This is going to be a little rough, typing the answer.  Here it goes:  \n(x-2)2(squared)+ (y-4)2 (squared)= 25\n\nYou find the midpoint between the two points for the center.  You find the distance between the two points for the diameter and divide that in half for the radius which equals 5.  The center is (2,4).  You square the radius and that is what comes after the equals sign.\n\nThe guy who entered before me is wrong.  I promise.
383381
The smallest cylinder that will contain the sphere has to be exactly as big around as the sphere and just as tall.  So, if the sphere has radius r, then the cylinder has radius r and height 2r.\n\nCorrect Volume of Sphere = 4/3 * (pi) * (r^3)\n\nVolume of Cylinder = (pi) * (r^2) * (2r)\n\nRatio of sphere to cylinder:\n\n(4/3 * (pi) * (r^3)) / ((pi) * (r^2) * (2r))\n\n= 2/3
237112
the cylinder shape and the contents.
954885
First lets assign the direction of f(1) as being at 0 degrees, then f(2) is at 57 degrees.\n\nResolve both forces in x, and y\n\nF(x) = f(1) + f(2) cos (57)\n\nF(y) = f(2) sin (57)             f(1) is at 0 degrees so has no y component.\n\nThe resultant force F = sqrt ( F(x)^2 + F(x)^2)\n\nThe angle of the resultant force \n\nArctan( F(y)/ F(x))\n\nYou do the number crunching
970605
For your first problem...\n\nYou know that T = Fc = (mv^2) / r\n\nWhere T is the Tension, Fc is the Centripetal Force, m is the mass, v is the velocity, and r is the radius.\n\nYou're given m = 80 kg, r = 0.90 m, and T = 600N.\n\nSo...\n\n600 N = (80kg * v^2) / (0.90 m)\nSolve for v...\nv = +/- 2.598 m/s\n\nSo since -2.598 m/s doesn't make sense (a negative speed), the solution would be ~ 2.6 m/s.\n\nFor your second problem...\n\nIts been a long time since I've done circular motion.  This site has the equation you need to use (its too messy to type here): http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html if you can figure out the angle.  So if you know what the angle should be, that equation will get you the coefficient of friction.\n\nIf you can't figure out the angle, at least attempting to solve it using that equation and some reasonable angle should get you some points I'd imagine.\n\nIf I can think of the correct angle I'll post an edit to this.
866787
let the radius of the cylinder be x,  drawing a cylinder in a cone on the paper, we see 2 triangles - the cone & the triangle with it's base as the top flat surface of the cylinder. follow me?\n\nif u do, then let's assume the height of the second triangle to be y. y can be found in terms of x as:\n(since both triangles are similar)\n\nx:4 :: y:12  \ni.e. x/4 = y/12\n=> y=3x\nthus ht of cylinder = ht of cone - y\n=12-3x\n\nvolume of cylinder = pi*r2*h  (r-> radius, h-> height of cylinder)\nv= pi * x2 * (12-3x) = pi * (12x2 - 3x3)\n  (the second term being 12 *x*x - 3*x*x*x ....sorry, i don't have a math equation display software here)\n\npartialy differentiate above equation wth respect to  x, we then get\ndv/dx = pi * (24*x - 9*x*x)\n\nequating dv/dx to zero, (to find a value of x that makes the volume value either maxium or minimum. u can verify it's max by again differentiating dv/dx with respect to x, equating to zero & finding if the value solved for is positive or negative)\n\nanyways, we get\n\neliminating  pi,\n24*x - 9*x*x=0\n\n=> x=24/9 = 8/3= 2.67\n\nnow, simply substitute this value of x in the volume equatiion above to get\n v= pi * (12*2.67*2.67 - 3*2.67*2.67*2.67)\n= 89.372 or so.\n\n\nuh...i guess i only confused u up, hope it helped u atleast a bit.
116020
Volume = pi*r^2*h\n\nWhere pi = 3.14.159...\nr = raduis of cylinder in feet\nh = height of cylinder in feet
919521
Imagine the center of the triangle. Draw a line segment from it to a vertex, and then one from the center to a midpoint of a side. This makes a 30-60-90 triangle, so the line segment is (p/2) sqrt(3). This is a radius of the cylinder, so we multiply it by 2 to get the diameter.\n\nd = p sqrt(3)\n\nor, as the question wants it\np = d/sqrt(3) = d sqrt(3)/3
924339
a) y= 10^(tan (πθ))\n= e^(ln10*tan(πθ))\n= e^u where u = ln10*tan(πθ\ndy/dθ =dy/du * du/dθ\n= e^u * ln10*πsec²(πθ)\n= ln10*πsec²(πθ)*e^(ln10*tan(πθ))\n= ln10.πsec²(πθ)*10^(tan (πθ))\n\nb) y= √[tln(t^4)]\n= √[4t*ln(t)]\n= 2√[t*ln(t)]\n= 2u^½ where u = t*ln(t)\ndy/dt = dy/du * du/dt\n= u^-½ * (ln(t) + 1)\n= (ln(t) + 1)/√[tln(t)]\n\nc) y = arctan(arcsin (x^½))\n= arctan u where u = arcsin v where v = x^½\ndy/dx = dy /du * du/dv * dv/dx\n= 1/(1+ u²) * 1/√[1 - v²] * ½x^(-½)\n= 1/[2√(1 - x)*(1 + (arcsin (x^½))²)]\n\nd)xe^y = y - 1\ne^y + xe^y * dy/dx = dy/dx\nSo dy/dy - xe^y * dy/dx = e^y\nie dy/dx (1 - xe^y) = e^y\ndy/dx = e^y/ (1 - xe^y)\n= 1/(e^(-y) - x)
728936
1. obtain formulas for surface area and volume of cylinder\n2. let volume equal 1000 and solve this volume equation for h\n3. substitute the expression from previous step into surface area    equation to get equation with only r's\n4. differentiate surface area formula with respect to r set this equal to 0 and solve for r i got   r=10/3rd root(pi)\n5. substitute value for r into volume equation solve for h  i got h=10/(pi)^(2/3)\n\nincidentally, i used 2nd derivative test to conclude r is minimum\n\n\n\ni'll get back to u with more
270860
You will need to know the mass of the material inside.\n\n1.  Subtract that from the overall mass.  The remaining mass will be the steel cylinder itself.\n\n2.  Find the volume of the steel by multiplying the mass by the density factor.\n\n3.  Volume of the cylinder itself is:\n\n  [((r1^2) * pi)  - ((r2^2) * pi)] * L  = Vol of Cyl.\n\nWhere: \n\n   r1 is the outside radius \n    (known, this is half the measured diameter)\n   \n   r2 is the inside diameter (unknown)\n    \n   L  is the length of the cylinder\n\n   Vol of Cyl is the mass times the density\n\n\n4. Solve for r2 with the previous equation:\n\n  [((r1^2) * pi)  - ((r2^2) * pi)] * L  = Vol of Cyl. \n\n\n  r2 = SQRT {[((r1^2) * pi) - (Vol of Cyl * L)] / pi}\n\n5.  r1 - r2 is your cylinder wall thickness.
651714
Homework?\nOK, only a hint. Consider a cylinder made of the circle and with a height dh. What is the volume of this cylinder? \n\nNow displace the cylinder along the path over dh, etc.