Question

Find parametric equations for the tangent line at t=2 for the motion of a particle given by x(t)=8t^2+6, y(t)=8t^3.

For the line,
x(t)=
y(t)=

Answer 1

just replace t =2 into them to get the answer

Answer 2

for a parametric equation? I put them in and it said they were wrong. I got 38 and 64. Don't you need to use the derivative to find the slope?

Answer 3

You find the derivative
\[
\frac{y'(t)}{x'(t)}=\frac{24 t^2}{16 t}=\frac{3}{2}t\\

\]

Answer 4

Replace t by 2 to fine the slope of the tangent at the point (x(2),y(2))

Answer 5

The slope is 2 and the point (38,64}
x(t)= 38 +t-2= 36 +t
y(t)=64 + 2( t-2)=60+2t

Answer 6

(32,96)? or am I replacing t=2 for 3/2t?

Answer 7

Do you understand it now?

Answer 8

uh oh. i think i'm lost. I found the slope with the derivative. How do I plug that in to get the equation? I'm not understanding that part.

Answer 9

Do you understand that the slope is equal to 2 at the point (38.64)

Answer 10

(38,64)

Answer 11

ok, yeah. I get that now

Answer 12

to get a line of slope 2 to go through (38, 64)
(x(t),(y(t))=(38,64)+ (t-2, 2(t-2))=(38+t-2,64+2t -4)=(36 +t,60+2t)

Answer 13

For t=2

Answer 14

Sorry, I have to go. It should be clear now.

Answer 15

Yeah I understand it. Thanks.