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
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 =...
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 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?