Question

Find an equation for the tangent line to the curve at point defined by the given value of t.
x=3t^2+8, y=t^6, t=-2

Answer 1

You differentiate both x and y with respect to t, then plug in t=-2

Answer 2

i did that but somehow i am not getting the right answer

Answer 3

what did you get as the derivatives?

Answer 4

6t^5/6t=t^5/t

Answer 5

well do X first, alone

Answer 6

ok, it will be 6t

Answer 7

right so substitute t=-2

Answer 8

-12

Answer 9

now do the same for y

Answer 10

6t^5=6(-2)^5=6*-32=-192

Answer 11

so i get 16

Answer 12

hmm, so, what does your book have as the answer?

Answer 13

i have to write equation for the tangent line. So, i found my slope which is 16. For x and y do i evaluate the given equations at the value of t?

Answer 14

yes, now that you have the slope you only need the intercept

Answer 15

so if you put t=-2 into the original equations you get the point the curve passes through at t=-2. Now you have both a point and a slope, and that's enough to write the equation for a line.

Answer 16

ok thanx ktklown.:)

Answer 17

np :)