Question

Find the equation of the tangent to the curve x=e^x, y =(t-1)^2, (1,1)

Answer 1

Slope of tangent is dy/dx
Please check x =e^t

Answer 2

take partials right?

Answer 3

what does it mean
\[x=e^x\]?

Answer 4

\[x=e ^{t}, y=(t-1)^{2}\]

Answer 5

we got that the tangent is y=-2x+3 but im not sure what number to evaluate the derivative at

Answer 6

dx = e^t ; dy = 2(t-1)

Answer 7

Take partial derivatives
dx/dt = e^t
dy/dt = 2(t-1)
Slope = dy/dx = dy/dt / dx / dt = 2(t-1)/ e^t

Answer 8

ok i got that far

Answer 9

but what number do we evaluate that at?

Answer 10

e^x = 1

(t-1)^2 = 1

Answer 11

.... typos persist

Answer 12

(1,1) is given (x,y) is (1,1) or is it t = 1

Answer 13

its supposed to be t=0 im just not sure where that is coming from

Answer 14

t = 0

Answer 15

e^t = 1; when t=0
(t-1)^2 = 1 when t=0

Answer 16

plug in t=0 into the derivative

Answer 17

but those parameters wweren't given in the question. are we just supposed to assume when t=0?

Answer 18

(x=1,y=1) is given

Answer 19

dy/dx = dy/dt / dx / dt = 2(t-1)/ e^t
Find slope at t =0
slope = -2
y = mx +c
Plug i n this value
And then find c by plugging in (1,1) for (x,y)

Answer 20

x(t) = 1 when t=0
y(t) = 1 when t=0

Answer 21

y = -2x +2(1) + 1

Answer 22

im still unclear where t=0 is coming from

Answer 23

do you see the point that they give you? that (1,1) in the question?

Answer 24

yes

Answer 25

(x,y) is the point; which means that (x=1, y=1)

x = e^t = 1

y = (t-1)^1 = 1

Answer 26

^2 ...

Answer 27

nice

Answer 28

got it thanks so much!

Answer 29

youre welcome