Robert F. answered • 08/17/15

Tutor

5
(10)
A Retired Professor to Tutor Math and Physics

v(t)=(t

^{2},5t)Part 1

x component is a parabola with vertex at (0,0) and opening upward.

y component is a line from (0,0) with a slope of 5.

Part 2

x(t)=5cos(t)

y(t)=-5sin(t)

x

^{2}+y^{2}=25[cos^{2}(t)+sin^{2}(t)]=25(1)=25This is the equation of a circle with radius 5 and center at (0,0)

Part 3

Bearings are measured relative to North with positive angles in the clockwise direction.

cos(60)=1/2

sin(60)=(√3)/2

cos(315)=1/√2

sin(315)=-1/√2

sin(60)=(√3)/2

cos(315)=1/√2

sin(315)=-1/√2

A vector with length 500 and angle 60 relative to north has x,y coordinates of (500sin(60),500cos(60))

A vector with length 40 and angle 315 relative to north has x,y coordinates of (40sin(315),40cos(315))

(500sin(60),500cos(60))=(500(√3)/2,500/2)=(250√3,250)

(40sin(315),40cos(315))=(-40√2,40/√2)

Add the vectors to get a resultant vector with x,y coordinates of

(250√3-40√2,250+40/√2)=(221.716,278.284)

The speed is the magnitude of the resultant vector.

Speed=sqrt(278.284

^{2}+221.716^{2})=355.809The direction is obtained by observing that the tangent of the angle is 278.284/221.716=1.225

The angle is arctan(1.225)=50.77 degrees (a northeast direction)

A general term that fits the terms that are given is

a

_{0}=(-1)^{0}x^{0}/[2^{0}(0!)=1a

_{1}=(-1)^{1}x^{1}/[2^{1}(1!)=-x/2a

_{2}=(-1)^{2}x^{2}/[2^{2}(2!)=x^{2}/8a

_{i}=(-1)^{i}x^{i}/[2^{i}(i!)Double check the above to make sure that there are no errors.

cos(315)=1/√2

sin(315)=-1/√2