- #1
nns91
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Homework Statement
1. Given the line x=(5/3)x and the curve C: x= sqrt(1+y^2). Let S be the region bounded by the two graphs and the x-axis. The line and the curve intersect at point p.
a. Curve C is a part of the curve x^2-y^2= 1. Who that x^2-y^2=1 can be written as the polar equation: r^2=1/(cos^2(\theta)-sin^2(\theta))
b. Use the polar equation given in part a to set up an integral expression with respect to the polar angle theta that represents the area of S.
2. A particle starts at point A on the positive x-axis at time t=0 and travels along the curve from A to B to C to D. The coordinates of the particle's position (x(t),y(t)) are differentialbe functions of t where x'(t)= -9cos(pi*t/6)sin(pi*sqrt(t+1)/2) and y'(t)=dy/dt is not explicitly given. At time t=9, the particle reaches its final position at point D on the positive x-axis.
(Note: Point A is right of point D so the particle is moving LEFT)
a. At point C, is dy/dt positive ? At point C, is dx/dt positive ?
b. The slope of the curve is undefined at point B. At what time is the particle at point B ?
c. The line tangent to the curve at the point (x(8),y(8)) has equation y=(5/9)x-2. Find the velocity vector and the speed of the particle at this point.
d. How far apart are the points A and D of the particle ?
Homework Equations
A= 1/2 integral(r^2 dtheta)
The Attempt at a Solution
1.
a. How should I prove this ?
b. I know the formula for area but how do I find the interval in term of theta ?
2.
a. So there is a graph given, at point C, the particle is going down so I think dy/dt is negative and as well as dx/dt right ?
b. So slope is undefined so dy/dx is undefined which means dx/dt =0 right ? so just find t for dx/dt =0. Am I right ?
c. so slope is 5/9 which means (dy/dt)/(dx/dt)=5/9. Does that mean that dy/dt=5 and dx/dt=9 ?
d. How do I attack part d ?