CALC 2 HELP! URGENT :( Best answer awarded asap :)?
1: Just plug in θ = 5π/4 r = 10 + 5cos(5π/4) = 10 + 5/√2 2: x = rsinθ = (10 + 5/√2) × (−1/√2) y = rcosθ = (10 + 5/√2) × (−1/√2) 3: Set 10 + 5cosθ = 0 and solve.
Calc 2 Help! Best Answer Awarded ASAP :)?
polar coordinates x = r cos t y = r sin t 8 (r cos t) - 2 (r sin t) = 5 r ( 8 cos t - 2 sin t) = 5 r = 5/(8 cos t - 2 sin t) b) (r cos t)^2 - (r sin t)^2 = 2 r^2 cos^2 t - r^2 sin^2 t = 2 r^2(cos^2 t - sin^2 t) = 2 r^2(cos 2t) = 2 r^2 = 2 sec 2t r = sqrt (2 sec 2t)
Calc 2 Help! Parametric Equations! Best answer awarded asap :)?
x=9t+ln(t); y=5t-ln(t)=> dx=9+1/t dy=5-1/t => dy/dx=(5-1/t)/(9+1/t)=> dy/dx=(5t-1)/(9t+1) d^2y/dx^2= [(9t+1)(5)-(5t-1)9]/(9t+1)^2= 14/(9t+1)^2
CALC HELP! PLEASE. BEST ANSWER AWARDED ASAP! :)?
Work is the amount of force projected over a certain distance. In this problem, you can use the change in potential energy as an equivalent to the amount of work done. That is, to move 1 kg of water up 1 m in a 9.8 m/s/s gravitational field, it requires 9.8 J of work. So, in short, you want to establish how much mass is moved a certain distance. So consider the following: m = D*V (mass = density * volume) m = D*A*h (mass = density * Area of a Section * Height of a section) dm = D*A*dh (since D and A are constants) With that final equation, we have sliced the pool into thin "wafers" of the infinitely thin height dh, having the infinitely small mass dm. Now, the work required for moving each wafer of water up is: dW = 9.8 * dm * h (dW is the work, dm is the mass of the wafer, and h is the height it is moved upwards) dW = 9.8 * D*A*dh * h (substituting our mass-density-volume equivalence from above...) We can now integrate to get: W = 4.9 * D * A * h^2 Keeping in mind that h is measured from our reference point and is NOT just the depth of the pool, the integral is evaluated from -4.0 to -2.5 m (as the pool is 3 m tall and has 1.5 m of water in it, and we want to move the water 1 m above the edge). The Density is given, and the Area is the surface area of the circular pool, 1/2 * pi * 25 m^2 or 39.3 m^2. So our final equation will be: W = 4.9 * 1000 * 39.3 * (4^2 - 2.5^2) Or about 1.88 million joules.
Calc 2 Series Question! Please Help!! :( Best Answer Awarded ASAP?
The series converges. This is a geometric sequence...here: a (first term) = -8/12 Common Ratio = -(2/3) The sum of a convergent series is given by: S = a/(1-r) S = (-8/12)/(1-(-(2/3))) S = -0.4
Please help me answer these calc 3 questions ASAP?
I guess you mean "intersecting" where you typed "interesting." (A) L1: r1 = <5,3,1> - t <12,-9,3> ; L2: r2 = <3,0,7> + s <8,-6,2> = <3,0,7> + (2/3)s <12,-9,3>. The two lines are parallel. (B) L1: Two points on this line are (1,0,5) and (0,1,2), so it may be written r1 = <0,1,2> + t <1,-1,3>. L2: Two point on this line are (2,3,7) and (4,-1,14), so it may be written r2 = <2,3,7> + s <2,-4,7>. The slope vectors are not parallel, so the lines are not parallel. If we now set: <0,1,2> + t<1,-1,3> = <2,3,7> + s<2,-4,7>, we get t = 2+2s AND 1-t = 3-4s AND 2+3t = 7+7s. The solution to the first two of these is found by adding the first two equations together, yielding: 1 = 5-2s => s = 2, t = 6. These values fail to satisfy the third equation (2+3t=7+7s), so the two lines don't intersect; they are skew lines.
Calculus help asap! 10 points best answer!?
For most land animals, the relationship between leg width w and body length b follows an equation of the form w=cb^(3/2) for some constant c>0. Show that if b is large enough, w'(b)>1. Conclude that for larger animals, leg width (necessary for support) increases faster than body length. Why does this put a limitation on the size of land animals? mind=blown...please help, i would really really appreciate it!!
Please help me... Best answer awarded asap!?
So there is this guy that lives a few floors below me in the dorms. I hav talked to him like one time for ten seconds in person, but we r friends on facebook and we emailed for like 4 days, then exchanged numbers last night via email. So from about noon till 8 pm he kept texting me, and if I wold stop he would start it up again. I know this is all normal and such for two friends, i mean he was in no way hitting on me... but is facebook relationship ststus says engaged, so for some reason I feel like it's wierd. I mean we dont know eachother he is engaged and texts me all day. Should I feel wierd about casually texting him for so long as nothing more than friends, or is an engaged man not supposed to spend all day texting some girl he barley knows... I know it sounds stupid but what do ya think??
Calc 2 Help! Highest and Leftmost Point according to graph! Best answer awarded asap! :)?
(1) x=5te^(t); y=4te^(-t)=4t/e^(t) => dy/dx=4e^(-t)[1-t]/{5e^(t)[1+t ]} dy/dx=0=>t=1 d^2y/dx^2=-2(e^2)/5<0, when t=1. =>t=1 gives a maximum. y(1)=4/e x(1)=5e => the Local max.point=(5e, 4/e). t does not exist for x->-infinity t<0=>y<0 & t=0=>x=0 & y=0 => (5e, 4/e) is the leftmost highest point of the curve. (2) The horizontal asymptote is y=0. The vertical asymptote is x=0.
Calc 2 Help! Maclaurin Series and Integration! Best answer awarded asap :)?
In terms of the McLaurin series, the expansion of sin6x^2 being, (6x^2) - (6x^2)^3/3 + (6x^2)^5/5 - ... Integrating by each term, we have ∫sin(6x^2).dx = 2x^3 - (108/7).x^7 + ... + C The value of the integral over the given range then being 2 * (0.62)^3 - (108/7) * (0.62)^7 + ... ~ 0.476656 + 0.543334825 ~1.02 Dunno if my calculations are correct, but the method should be good.