Question

Parametric Equations... Help?

Answer 1

which number

Answer 2

52. fidn the 2nd derivative of that and to find c.u. set your 2nd deriv = to 0

Answer 3

50.) primarily...
it's hard haha @marcelie

Answer 4

lool hmm @Directrix

Answer 5

they've given you all the formulae for the first one

you just need to pattern-match the numbers in

Answer 6

k

Answer 7

you can see from chain rules
\[\huge \frac{ dy }{ dx }=\frac{ dy }{ dt }*\frac{ dt }{ dx }\]

Answer 8

x(t)= (vsubzero cos alpha)t
y(t)= (vsubzero sin alpha)t-(1/2gt^2)

Let g=32ft/sec^2
Let vsubzero = 60ft/sec^2
foot ball player kicks 36 ft. above ground

Answer 9

you can divide by dx/dt instead to be

\[\large \frac{ dy }{ dx }=\frac{ \frac{ dy }{ dt } }{ \frac{ dx }{ dt } }\]

Answer 10

@DanJS i got ya. are you doing 51?

Answer 11

yeah, that is the first derivative, the second would be easier to see

Answer 12

oh that is for 52, derivatives of the parametric equations

Answer 13

which one you want to do?

Answer 14

51.)
Particle A
x(t)=4t-4
y(t)=2t-k
dx/dt = 4
dy/dt=2

Particle B
x(t)=3t
y(t)=t^2-2t-1
dx/dt=3
dy/dt=2t-2

a.) if k=5 do particles ever collide? particles only collide when x,y particle a = x,y particle b.
so:
x(t)A=x(t)B
4t-4=3t
so when t=4 the particles collide in y direction...
plug into y
y(4)=4^2-2(4)-1 = 7
y(4)=8-k
7=8-k
k=1
------
Particles don't collide when k=5

Answer 15

@DanJS
50.) is the only that is tripping me up haha

Answer 16

I don't know how to write the x and y component?

Answer 17

x and y components of velocity?

Answer 18

hi

Answer 19

@imqwerty ... really confused on how to do 50a.)

Answer 20

@DanJS hi!

Answer 21

for part a) they give you the parametric equations for the position of a kicked ball in gravity.

x(t) and y(t) are given, that tells you the x and y values for the location of the ball for some time t.

Answer 22

does alpha change? or is it always 36 degrees

Answer 23

Call the posistion where the ball starts is the origin (x,y)=(0,0), the position t seconds after the kick is
\[x(t)=(v _{0}\cos(\alpha))*t\]
\[y(t)=(v _{0}\sin(\alpha))t-\frac{ 1 }{ 2 }g*t^2\]

Answer 24

for 50c.) does the football hit the ground when dy/dt = 0? essentially a straight line?

Answer 25

they give you the original velocity , v0=60 ft/s, and the angle of the kick, alpha=36, and the gravitation constant g=-32 ft/s^2

just put those in the parametric equations and simplify

Answer 26

k. i understand d... you just find dx/dt and dy/dt @ t= 1 and solve for the speed using the Pythagoras theorem.

Answer 27

x(t) = 48.54*t
y(t) = 35.27t + 16t^2

Answer 28

thank you!

Answer 29

To tell when the ball is at the ground level, you set the vertical position to 0, y(t)=0

y(t) = 0 = 35.27t + 16t^2

Answer 30

ohhh okay!!!

Answer 31

sorry should be a minus

y(t) = 0 = 35.27t - 16t^2

solving gives time t=2.2 sec

Answer 32

right. thank you so much. and when the ball is at it's highest point dy/dt=0... and to find the maximum height you just plug that value you t back into y(t)