Question

Pls help

Find an equation of the normal line to the graph of xe^xy = 7 at the point where x = 7

:) thank you :)

Answer 1

is the equation like this? \[\Large xe^{(xy)}=7\]

Answer 2

should we assume the y is a function of x and do an implicit derivative?

Answer 3

once you know the tangent line, you can always flip it around to a normal

Answer 4

yes @sirm3d

Answer 5

implicit derivative please @amistre64

Answer 6

i take it you might be confused about how to operate implicit stuff then?

Answer 7

yes :(

Answer 8

hmm, do you know how to work the chain rule, the product rule, and the e^x rule ?

Answer 9

yup

Answer 10

then thats all there is to this problem, is just the name of the variables that is messing with your mind :)

Answer 11

x e^(xy) = 7; lets clean it up by using a and b

ab = 7; this is just a product rule

ab'+a'b = 0 ; since a=x in this case, and b=the e parts ....

xb'+x' e^(xy) = 0 ; and x' = 1 in this case

xb'+e^(xy) = 0 all thats left is to determine the b'

Answer 12

lets define b as: e^(pq) just to make it general

we know e^f(x) = f'(x) e^f(x) right?

Answer 13

the book might have it as: e^u derives to u' e^u

Answer 14

so it will wnd up as y=1n1/7 ?

Answer 15

since b=e^(pq); then b' is (pq)' e^(pq)

(pq)' is just the product rule again: pq' + p'q,

maybe, i havent seen the future yet :)

Answer 16

i followed what you did.. and i think i got that :) hehe

Answer 17

how do i get the equation of a normal line ?

Answer 18

good :)

x e^(xy)

x (xy'+y) e^(xy) + e^(xy) = 0; and x=7

7 (7y'+y) e^(7y) + e^(7y) = 0

e^(7y) (49y'+7y +1) = 0

is there a value for y that you might have missed? or maybe we solve that in the original.

the key is getting to y' and defining the tangent line at the given point

Answer 19

once we know the tangent line: y-yo = m(x-xo)

the normal line is just: y-yo = -1/m (x-xo)

Answer 20

7 e^7y = 7

e^7y = 1

y = 0

Answer 21

well that neatens this up alot lol

Answer 22

e^(7y) (49y'+7y +1) = 0 ; y=0

e^(0) (49y'+0 +1) = 0

49y'+1 = 0

y' = -1/49

Answer 23

y' is the slope of the tangent; therefore the perp to it is just 49

Answer 24

so, just construct the normal line using a slope of 49 and a given point of (7,0)

Answer 25

yayy i got it !!!

y= 49x-343 ?

Answer 26

since the negative reciprocal of -1/49 is the slope and is 49 ?

Answer 27

or no... i read it from somewhere ?

Answer 28

that looks pretty good to me

Answer 29

thank youuu!!!

Answer 30

http://www.wolframalpha.com/input/?i=xe%5Exy+%3D+7%2C+y%3D+49x-343%2C+y%3D-%28x-7%29%2F49

the picture it shows is not very good, but it looks like it all works out ;)

good luck