differentiation

[grim moon]

i cant do b

[charred dagger]

do you know what a normal is

[grim moon]

yeah

so dy/dx is = 0

thats what i know

[charred dagger]

knowing that dy/dx = 0 gives you an equation of x and y

keep in mind though this normal needs to still be on the curve

so x^5 + y^5 = 5x^2 y is a second equation of x and y

as a fraction (2xy - x^4)/(y^4 - x^2) = 0 if 2xy - x^4 = 0 and y^4 - x^2 ≠ 0

so you have the two equations

• x^5 + y^5 = 5x^2 y
• 2xy - x^4 = 0
  to solve for x and y

you can try solving for x and y now

once you get an x and y, the normal has to go through the curve at x = h
therefore the answer is the x you got

[grim moon]

thanks i will try

can i check my answer with you?

[charred dagger]

sure

[charred dagger]

however you can just check your answer by plugging in the values you get back into the two equations

you only need to find an x and a y that fit both equations
once you find them, youre good

[grim moon]

i will plug it in but for now i think h = 2.52

or cube root of 16

hm its wrong i think

[charred dagger]

yea its wrong unfortunately

to speed up your checking, Ill tell you that h is between 1 and 2

they tell you that h > 0 in the problem, this is enough to limit your solutions down to just one value

in other words x > 0

[grim moon]

maybe my equation for y in terms of x is wrong then

cuz i got y=(x^3)/2

[charred dagger]

thats correct

try that again then

[grim moon]

no i have just no idea how im not getting a right answer

[charred dagger]

(x^3)^5 = x^3 * x^3 * x^3 * x^3 * x^3 = x^15

[grim moon]

oh my god

im stupid

than you sm

so its the 10th root of 16

but that still isnt right

hm

[charred dagger]

redo how you factored

[grim moon]

ah

3*16

ok its actually right now

thanks

how do i mark this as solved?

[charred dagger]

do .solved

[grim moon]

.solved