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November 27, 2014

November 27, 2014

Posted by **bryan** on Sunday, February 3, 2013 at 12:57pm.

- Calculus -
**Damon**, Sunday, February 3, 2013 at 1:15pmA is angle in radians between perpendicular to shore at P (length 3 ) and light at shore (x which is 1 at start).

tan A = x/3

dA/dt = w = rad/sec = 4* 2 pi/60 = .419 rad/s

dx/dt = 3 d tan A /dt

dx/dt = (3/sec^2 A) dA/dt

dx/dt = 1.26 /sec^2 A = 1.26 cos^2 A

at x = 1, tan A = 1/3

so A = 18,43 deg

so cos A = .949

1.26 cos^2 A = 1.13 km/s

- Calculus -
**H H Chau**, Friday, August 22, 2014 at 4:14pmConsider a right angled triangle,

tan(A)=x/3

cos(A)=3/sqrt(x^2+9)

sec^2(A)=(x^2+9)/3^2

dA/dt=4 rev/min = 2*pi/15 rad/s

Also,

x=3 tan(A)

dx/dt=3 sec^2(A) dA/dt

dx/dt=(x^2+9)/3 * 0.419

At x=1, dx/dt=1.396 km/s or 5027 km/hr.

The above attempt confused sec^2(A) with 1/sec^2(A), and 1/cos^2(A) with cos^2(A).

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