Question

r(t)=(8cost)i + (-6t)j find vector equation of the line which is tangent to its graph at the point P0=(2,-PI)

Answer 1

r'(t) =(-8sint) i -6 j
to find it at P0=(2,-PI) find t for which 8cost = 2 and -6t = -PI
so cost =1/4 or t=Pi/6 which doesn't look possible....:) So I think there is a mistake...
Forgeting about this , imgine value of t is consistent, then just put this value in the expression of r'(t) and write the line equation in the form (x,y) =(2,-PI) +T r'(t) where 0<T<infinity

Answer 2

lets assume that\[r(t)= (4\cos t)i +(-3t)j\] and the points are \[P0(2,-pi)\]

Answer 3

the thing is that the point P0=(2,-Pi) has to belong to the graph r(t)=(4cost)i + (-6t)j
so: 4cost=2 and -6t=-Pi has a solution. In this case t=Pi/6 and cost = 1/2 which is not true.

Answer 4

Ah sry i dind't see -3t.......
Then it would be ok. Value of t = Pi/3
so r'(Pi/3)=-4sqrt3 i -6j
so tangent line is: (x,y) =(2,-PI) + T(-4sqrt3,-6) where 0<T<infinity

Answer 5

ok?

Answer 6

there did the 6 come from?

Answer 7

ahh sry you change it to -3
r(t)=(8cost)i + (-3t)j
r'(t) =(-8sint) i -3 j

Answer 8

(x,y) =(2,-PI) + T(-4sqrt3,-3) where 0<T<infinity

Answer 9

what if the equation was r(t)=(4cost)i+(−3t)j

Answer 10

just folow the steps i said you

Answer 11

1º find value of t for the point
2ºfind the r'(t) in this t
3º write the equation in vector form (x,y) = p0 + Tr'(t)

Answer 12

1º 4cost= 2, -3t=-Pi
so t=Pi/3 and 4cosPi/3 = 1/2 .Everything it's ok...
2º r'(t) = (-4sint)i +(-3)j.........r'(pi/3)=-4sqrt3/2 i -3j
3º (x,y)= (2,-Pi) +T(-2sqtr3,-3)

Answer 13

give medal or say thankyou....

Answer 14

thanks for the help

Answer 15

you welcome

Answer 16

@myko please do not demand reparations. that is up to the prerogative of the asker.