Question

Find the equation of the tangent to the curve defined by y =e^x that is
perpendicular to the line defined by 3x+y=1.

Answer 1

There are three pieces of information here:
1. it's a straight line (any tangent must be a straight line)
2. it's tangent to the curve y=e^x
3. it's perpendicular to the line 3x+y=1

Answer 2

I got y=1/3 e^x but the answer should be
x-3y+(1+ln3)=0
How do I get that answer?

Answer 3

Let's find the slope of the line from the third piece of information

Answer 4

m=-3 in the line so slope of perpendicular line should be 1/3

Answer 5

yep

Answer 6

At which point of the curve y=e^x will the slope of the tangent be 1/3?

Answer 7

um -1?

Answer 8

Why so?

Answer 9

ln1/3?

Answer 10

I'm asking for a point... A point has 2 coordinates...

Answer 11

(-1,1)

Answer 12

where did the ln1/3 go

Answer 13

this point ain't even on the curve

Answer 14

lol idk o.o

Answer 15

At which point of the curve y=e^x will the slope of the tangent be 1/3?

Answer 16

ln1/3 is near

Answer 17

I'm not sure.

Answer 18

3e?

Answer 19

nope

Answer 20

never mind

Answer 21

On the curve y=e^x, when x=ln1/3, y=?

Answer 22

1/3

Answer 23

now you can answer

Answer 24

Now you know that the slope of the line is 1/3, and the line passes through the point (ln1/3,1/3)

Answer 25

And you should be able to find the equation of the line

Answer 26

I got x-3y-ln1/3 +1=0

Answer 27

Mind showing us the steps?

Answer 28

y=1/3 (x-ln1/3) + 1/3
expanded and multiplied whole thing by 3

Answer 29

The answers are all equivalent

Answer 30

How so?