Recent content by evolution685

sweet. that saved me several hours of pounding my head.

first i had to show solve x′=sin(x) to get t=ln|(csc(x₀)+cot(x₀))/(csc(x)+cot(x))| i did that. next i need to show that for x₀=(π/4) you can solve x=2arctan(((e^{t})/(1+√2))) what I've done so far is t=ln|((csc(pi/4)+cot(pi/4))/(csc(x)+cot(x))| t=ln|((2/sqrt(2)+1)/(csc(x)+cot(x))|...

my next problem is that it says show that for x0=pi/4 you can solve x=2arctan((e^t)/(1+sqrt(2)) here's what i have so far, working from the previous equation: t=ln|((csc(pi/4)+cot(pi/4))/(csc(x)+cot(x))| t=ln|((2/sqrt(2)+1)/(csc(x)+cot(x))| e^t=(2/sqrt(2)+1)/(csc(x)+cot(x))...

ahhhhh. i see. thank you.

i'm having trouble with this. i first need to solve x'=sinx. the answer is given as t=ln|(csc(x0)+cot(x0))/(csc(x)+cot(x))|] i'm not sure where those x0's came in from. here's what i did: Int(1/sinx)dx=Int(1)dt Int(csc(x))dx=Int(1)dt...

ok, i just replaced 200 with x-200 and 100 with x-100 in those equations then resolved. answer came out to 11am. i guess that makes sense because it's a linear equation so snow started at 11am, by 12pm 300ft were cleared, by 1pm 200ft were cleared, by 2pm 100 ft were cleared.

thanks for the reply. here's what i got: (S=integral) s'=x-300 Sds=Sx-300dt (1) s=t(x-300)+c 200=1(x-300)+c 500=x+c (2) c=500-x 100=2(x-300)+c (3) 700=2x+c (2) into (3) 700=2x+(500-x) x=200 700=2(200)+c c=300 (1) with constants added s=t(200-300)+300 s=100t+300...

ok so the equation would be dS/dt=x-300 where x is the amount of snow falling? then integrating would yield S=t(x-300). but I'm only given information about snow removed, not snow added (x) so I'm still confused.

i don't follow. what I'm having trouble with is formulating the actual equation. would it be dS/dt = S - 300? if that's not correct what would it be? and if it is correct then how would i go about separating it? i can't figure out how to get the dS and S on one side and the dt on the other...

Homework Statement A snow blower clears 300 cubic ft of snow per hour regardless of depth of snow. At some point before noon, it starts snowing and snows steadily all day. A man starts clearing a sidewalk with the snow blower at noon. He clears 200 feet in the first hour, and 100 feet in the...