Question

find an equation of the tangent line to the curve at the given point x=1+sqrt(t), y=e^t^2 (2,e)

Answer 1

tangent line given point (x1, y1)
\[y - y_1 = f'(x_1) (x - x_1)\]

get y in terms of x
take derivative

Answer 2

i took the derivative and this was my set up...is it right?\[\frac{ e ^{2t} }{ 2\sqrt{t} }\]

Answer 3

you need dy/dx , sub out the "t" variable first

Answer 4

dy/dx = (dy/dt) / (dx/dt),

Answer 5

that works too

Answer 6

\[dyd/x = \frac{2t e^{t^2}}{\frac{1}{2}t^{\frac{-1}{2}}}\]

Answer 7

that's supposed to be *dy/dx*

Answer 8

plug in t = 1

Answer 9

y - e = m (x - 2)

whatever the value of m is after you plug in 1 for t

Answer 10

is it 0 when you plug in 1 to the top equation? 2(1)e^(1)^2

Answer 11

no it's 2e

Answer 12

so m = 2e / (1/2) = 4e

Answer 13

is the answer y=4ex-9e?

Answer 14

y=4ex-7e