Recent content by aub

i guess i got it just tell me if its right so for the area of the inner loop A1, α = 2∏/3 and β = 4∏/3 and i apply the formula for the area between the inner loop, i find the whole area with α = 0 and β = 2π then substract A1

also, if anyone can help me to find the area of the region R between the inner loop and outer loop what would be α and β ?

yeah, i thought what i need to do is solving for θ, it gave me α = 2∏/3 and β = 4∏/3 and i here i don't need to multiply the integral by 2 anymore.. I am lost :mad: do you mean the symmetry?

i did the polar one, i can see that its tangent to 2∏/3 on the origin but i can't get the point if there is any and for the cartesian one, we never did that in class for those curves.. what were you hinting about? thanks

Homework Statement Find the area inside the inner loop of the limacon curve : r = 1 + 2cos(θ) Homework Equations A = ∫\stackrel{α}{β}(\frac{1}{2}r2)dθ The Attempt at a Solution i have the solution, my question is : how do you find α and β ? here α = 2π/3 and β = π A =...

ok thanks i solved it

i know this doesn't fit here right but i didnt know where else i could ask this Homework Statement how did they solve 1 through 4? it should be easy but i can't solve it :frown: thanks for any help

my answer my be right because i just found out that there are many changes in newer and older versions and that the book has apparently used other close values (like in many other exercices) so it may be that is what I am doing right?

Homework Statement If a wave y(x,t) = 5mm sin(kx + (600rad/s)t + φ) travels along a string, how much time does any given point on the string take to move between displacements y = 2mm and y = -2mm The Attempt at a Solution we have 1. 2=5sin(kx + 600t1 + φ) sin(kx + 600t + φ)=0.4...

Homework Statement The p-V diagram in the figure below shows two paths along which a sample of gas can be taken from state a to state b, where Vb = 4.0V1. Path 1 requires that energy equal to 3.0p1V1 be transferred to the gas as heat. Path 2 requires that energy equal to 8.5p1V1 be...

its 22mm is 0.022m not 0.22

its one formula that I am using and i already did give it (ΔL = α*Li*ΔT) So ΔL[ruler]= α[steel]LΔT= 11*10-6*20.11*230=-0.0508783 ΔL[rod]= -ΔL[ruler]+20.11-20.05= 9.1217*10-3 α[rod]=ΔL[rod]/[L*ΔT]= 1.9*10-6

ΔL = αLiΔT ΔL= change in length α= coefficient of linear expansion ΔT= change in temperature So ΔL[ruler]= α[steel]LΔT= 11*10^-6*20.11*230=-0.0508783 ΔL[rod]= -ΔL[ruler]+20.11-20.05= 9.1217*10^-3 α[rod]=ΔL[rod]/[L*ΔT]= 1.9*10^-6 right?

ΔL(steel)= αLiΔT= 11*10^-6*20.05*230= 0.0507 cm i didnt get your point though..

i found α for steel but i got stuck with the measurements since the ruler's 1 cm is now bigger any help? thanks