Question

Proof For 0 < t < pi/2

2sin(t) + tan(t) >3t

Answer #1

Find the length of the curve
1) x=2sin t+2t, y=2cos t, 0≤t≤pi
2) x=6 cos t, y=6 sin t, 0≤t≤pi
3) x=7sin t- 7t cos t, y=7cos t+ 7 t sin t, 0≤t≤pi/4

tan(t)=1/tan(t) in the interval 0≤t≤2π

Find all theta from (-pi,pi) for which the tangent line of
r=tan(theta/2) is horizontal.
I have gotten to
1/2sec^2(theta/2)*sin(theta)+tan(theta/2)*cos(theta)=0
How do I solve for 0?

Solve the below boundary value equation
1. Ut=2uxx o<x<pi 0<t
2. u(0,t) = ux(pi,t) 0<t
3. u(x,0) = 1-2x 0<x<pi

Given the function f(t) = 1/2cos (pi/3 *t) +3pi^2. Which of the
following is equal to f'(t)
f'(t) = 1/2sin(pi/3 t) +6pi
f'(t) = -pi/3cos(pi/3 *t)
f'(t) = -pi/6 sin(pi/3 T)
f'(t) = pi/3 sin(pi/3 t) +6pit

Find the length of the curve x= (3t - t)^3, y = 3(t^2) for 0 ≤ t
≤ 2

Solve for y(t) for y'+2/(120+3t)y=15 when y(0)=0

Find the curvature of the space curve x=(3t^2)-t^3 y=3t^2
z=(3t)+t^3

Solve the wave equation:
utt = c2uxx, 0<x<pi, t>0
u(0,t)=0, u(pi,t)=0, t>0
u(x,0) = sinx, ut(x,0) = sin2x, 0<x<pi

Find the length of the curve x = 3t^(2), y = 2t^(3) , 0 ≤ t ≤
1

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