Question

Find an equation of the tangent to the curve at the point by first eliminating the parameter

Answer 1

x=e^t , y=(t-6)^2 ; (1,36)

Answer 2

I tried to elem. the parameter and got lnx^2 - 12lnx +36 ?

Answer 3

why would we want to eliminate the parameter? just because it says so?

Answer 4

Correct.

Answer 5

Did I even do it right?

Answer 6

ln(x) = t is suitable sure

Answer 7

Hum. So what do I do to get the equation? I cant just plug in (1,36) for x and y because ln(1) is 0.

Answer 8

well, since y = (t-6)^2, and t=ln(x)

y = (ln(x)-6)^2

Answer 9

in order to make a line, we need a point and a slope ... how do we find the slope?

Answer 10

or lnx^2-12lnx+36

Answer 11

no need to expand it ..

Answer 12

Hum. Ok. So y-y1=m(x-x1)

Answer 13

good, and we have y1 and x2, we only need m

Answer 14

typos lol

Answer 15

So just take delta y/delta x? Which is ... um...

Answer 16

which is chain rule ...

Answer 17

Oh. Duh. Gotta take deriv. of that.

Answer 18

yep

Answer 19

Give me a sec!

Answer 20

2(lnx-6)*1/x

Answer 21

very good, and when x=1? m=?

Answer 22

-6?

Answer 23

2(0-6)/1 is not -6

Answer 24

It is if you are lazy and plug it in your calc wrong. :)

Answer 25

lol

Answer 26

-12

Answer 27

so whats our line?

Answer 28

y=-12x+

Answer 29

gott get that last number.

Answer 30

y-36 = -12(x-1) is the equation of a line ... but format however you want

Answer 31

Okie dokie.

So just -12x+1+36?
So y=-12x+37?

Answer 32

-12x -12(-1) +36

so no

Answer 33

Story of my life.
How about -12x+12+36
or y=-12x+48

Answer 34

looking good lol

Answer 35

So algebra is harder than calculus. Thank you.

Answer 36

i agree :)