F1=<10,0> and F2=5<cos(theta), sin(theta)>
a.
find ||F1+F2|| as function of theta
First we find the component of F1+F2
F1+F2=<10,0>+5<cos(theta), sin(theta)>
F1+F2=<10,0>+<5cos(theta), 5sin(theta)>
F1+F2=<10+5cos(theta),0+ 5sin(theta)>
F1+F2=<10+5cos(theta), 5sin(theta)>
Now, for magnitude of F1+F2
||F1+F2|sqrt((10+5cos(theta))^2+(5sin(theta))^2 )
||F1+F2|sqrt(100+100cos(theta)+25cos^2(theta)+25sin^2(theta) )
||F1+F2|sqrt(100+100cos(theta)+25(cos^2(theta)+sin^2(theta) ))...using cos^2(theta)+sin^2(theta)=1
||F1+F2||=sqrt(100+100cos(theta)+25)
||F1+F2||=sqrt(125+100cos(theta))
||F1+F2||=sqrt(25(5+4cos(theta)))
||F1+F2||=5sqrt(5+4cos(theta))
b.
Given
F1=<10,0>
F2=5<cos(theta), sin(theta)>
We have to Use a graphing utility to graph the function in part for a. for 0 <=theta <=2pi.
The graphing utility to graph the function is shown below.
c.
form part b range for 0<=θ<2 π
.
||F1+F2||=5sqrt(5+4cos(θ))
From here we get the maximum value of θ at θ = 0 °
||F1+F2||=5sqrt(5+4cos(0))
||F1+F2||=5sqrt(5+4*1)
||F1+F2||=5sqrt(9)
||F1+F2||=5*3
||F1+F2||=15
Now we get the minimum value of θ at θ = pi
||F1+F2||=5sqrt(5+4cos(pi))
||F1+F2||=5sqrt(5+4(-1))
||F1+F2||=5sqrt(5-4)
||F1+F2||=5sqrt(1)
||F1+F2||=5
The range of the function for 5< =θ<15 .
d. The magnitude of a vector is always a positive number or zero it cannot be a negative number. A vector is a quantity described by a magnitude and a direction. The magnitude is always positive or zero. A negative sign in front of a vector indicates the same magnitude but in the opposite direction. The negative sign is part of the direction rather than the magnitude.
