Question

y^4 + x^3 = y^2 + 10x, find the line that is normal to the curve at (0, 1)

Answer 1

what was your dy/dx ?

Answer 2

I ended up getting y' = (-3x^2 + 10)/y, but I have no idea if that is correct or where to go from there. ><;

Answer 3

it should be (-3x^2 + 10)/(4y^3 +2y)

Answer 4

Oh! Right! Forgot to apply the power rule to my y.

Answer 5

so, what is dy/dx at (0,1) ?

Answer 6

Would I just plug in the x and y values? Cause if so, I got 5/3 from doing that. Is that the slope?

Answer 7

right.. 5/3 is slope for the tangent.
so, what will be the slope of the normal ?

Answer 8

Oh! -3/5, yes?

Answer 9

correct.
now for that line.. -3/5 is the slope, and it passes through (0,1) .. you can easily find the eqn now

Answer 10

Is it y = (-3/5)x + 1?

Answer 11

nop

(y-0) = (-3/5)(x-1)

Answer 12

Oh, so just y = (-3/5)x then?

Answer 13

its x-1 pn RHS

Answer 14

i mean separately in brackets, final ans is not that

Answer 15

Pffft. I am stumped then. xD

Answer 16

y = (-3/5)(x-1)

whats the problem in simplifying RHS ?

Answer 17

To me it seems it should simplify to (-3/5)x + 3/5

Answer 18

that is correct.

Answer 19

if you want to simplify more, you can multiply both sides by 5

Answer 20

So it could be 5y = (-3)x + 3?

Answer 21

right

Answer 22

Ah, okay. Thank you very much for your help!

Answer 23

glad to help