Implicit Differentiation question

Ok first time here so hi,

we got this question on a hwk l8ly (studying P3)

Find d²y/dx² in terms of x if cos y + sin y = x

Many of us dont have a clue, any offers?

Reply 1

k@tie

b00

Ok first time here so hi,

we got this question on a hwk l8ly (studying P3)

Find d²y/dx² in terms of x if cos y + sin y = x

Many of us dont have a clue, any offers?

I'll give it a go...
so differentiating w.r.t.x:-
-sin y .dy/dx + cos y .dy/dx = 1
dy/dx (cos y - siny) = 1
dy/dx = (cos y - sin y)^-1
And differentiating again let u=cos y - sin y ==> du/dx = (-sin y - cos y).dy/dx
dy/du = -u^-2 = -(cos y - sin y)^-2
d^2y/dx^2 = -(cos y - sin y)^-2 . (-sin y - cos y).dy/dx
= ((sin y + cos y)dy/dx)/(cos y - sin y)^2
substituing dy/dx gives
(sin y + cos y)/(cos y - siny)^3 = x/(cos y - siny)^3
So somethings gone wrong here coz I don't have the denominator in terms of x....but thats the general idea...anyway! sorry!!

Reply 2

Fermat

b00

Ok first time here so hi,

we got this question on a hwk l8ly (studying P3)

Find d²y/dx² in terms of x if cos y + sin y = x

Many of us dont have a clue, any offers?

cos y + sin y = x
cosy.sin(pi/4) + siny.cos(pi/4) = x/rt(2) (multiply every term by cos(pi/4) = sin(pi/4) = 1/rt(2) )
sin(y+pi/4) = x/rt(2)

differentiate wrt x

cos(y+pi/4).y' = 1/rt(2)

differentiate wrt x, again

cos(y+pi/4).y'' - sin(y+pi/4).y'.y' = 0
y'' = tan(y+pi/4).(y')²
y'' = tan(y+pi/4) / [2cos²(y+pi/4)]
y'' = sin(y+pi/4) / [2cos³(y+pi/4)]

the next few lines are more easily followed if you write them out.

y'' = x/rt(2) / [2(1-x²/2)^(3/2)]
y'' = x / [2rt(2)(2-x²)^(3/2) / (rt(2))³]
y'' = x / (2-x²)^(3/2)
================

Edit: fixed arithmetic error - courtesy of Jonny :smile:

Reply 3

Jonny W

Suppose y is in (-3pi/4, pi/4).

sqrt(2) sin(y + pi/4) = x
y = arcsin(x/sqrt(2)) - pi/4
dy/dx = 1/sqrt(2 - x^2)
d^2y/dx^2 = x / (2 - x^2)^(3/2)

Suppose y is in (pi/4, 5pi/4).

sqrt(2) sin(y + pi/4) = x
sqrt(2) sin(y - 3pi/4) = -x
y = -arcsin(x/sqrt(2)) + 3pi/4
dy/dx = -1/sqrt(2 - x^2)
d^2y/dx^2 = -x / (2 - x^2)^(3/2)

Continuing the pattern, if y is in (5pi/4, 9pi/4) then

d^2y/dx^2 = x / (2 - x^2)^(3/2)

and if y is in (9pi/4, 13pi/4) then

d^2y/dx^2 = -x / (2 - x^2)^(3/2),

etc.

Reply 4

b00

chrs m8s