SOLUTION: 1)A particle moves along the top of the parabola y^2=2x from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the poin

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 1)A particle moves along the top of the parabola y^2=2x from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the poin      Log On


   

Question 846542: 1)A particle moves along the top of the parabola y^2=2x from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the point (2,2).
2)The quarterback of a football team releases a pass at a height of 7 feet above the playing field, and the football is caught by a receiver 30 yards directly downfield at a height of 4 feet. The pass is released at an angle of 350 with the horizontal.
Find the speed of the football when it is released.
Find the maximum height of the football.
Find the time the receiver has to reach the proper position after the quarterback releases the football.
Question 3
Evaluate the double integral over the given region R:
∬▒( xy^3)/(x^(2 )+ 1) dA,R:0≤x≤1,0≤y≤2
∬▒〖xye^(〖xy〗^2 ) 〗 dA,R:0≤x≤2,0≤y≤1
∬▒〖y sin⁡〖(x+y)〗 〗 dA,R: -π≤x≤0,0≤y≤π

5)Find the volume of the region bounded above the paraboloid z= x^2+ y^2 and below by the triangle enclosed by the lines y=x,x=0 and x+y=2 in the xy plane.
6)Find the area of the region that lies inside the cardioid r=1+cos⁡θ and outside the circle r=1.



Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
1) Velocity of the particle along parabola is in the direction of its tangent.
dy%2Fdx+=+1%2Fy+=+1%2Fsqrt%282x%29

At (2,2), dy%2Fdx+=+1%2Fsqrt%282%2A2%29=1%2F2.

Let the velocity be represented by: V+=+5cos%28theta%29i+%2B+5sin%28theta%29j


tan%28theta%29=dy%2Fdx+=+1%2F2

cos%28theta%29=2%2Fsqrt%285%29

sin%28theta%29+=+sqrt%281+-+cos%5E2%28theta%29%29+=+sqrt%281+-+4%2F5%29+=+1%2Fsqrt%285%29

Thus, the velocity vector is:
V+=+5%28%282%2Fsqrt%285%29%29i+%2B+%281%2Fsqrt%285%29%29j%29
V+=+2%2Asqrt%285%29i+%2B+sqrt%285%29j%29
If you want help with the remaining questions then send me an email. I can solve them on paper, scan and send you. Cheers!!!