Question

Find (d^2y)/dx^2 if x^2+y^2=25 at x=4.

Answer 1

@lizlozada did you try this?

Answer 2

try what?

Answer 3

this question

Answer 4

yes, but I'm stuck

Answer 5

ok, Would you mind sharing your work here?
I'll guide you then

Answer 6

2x+2yy'=0 (derivative of 25=0)

Answer 7

THEN

Answer 8

y'=-x/y what's next?

Answer 9

Now we need to take the second derivative
\[2x+2yy'=0\]
\[\frac{d}{dx}(2x+2yy')=0\]
\[2+2[y'\times \frac {d}{dx} (y)+2y y'']=0\]
would you simplify this?

Answer 10

why don't you take the derivative from -x/y?

Answer 11

We could do that also. you try this. Then I'll take this way also

Answer 12

i don't know how to do that though...

Answer 13

It's easy.
\[2+2[y'\times \frac {d}{dx} (y)+2y y'']=0\]
you just have to take derivative of y and plugin here
\[2+2[y'\times \large{\underline{\frac {d}{dx} (y)}}+2y y'']=0\]
then just simplify it

Answer 14

because when I asked my teacher, he said to take the derivative of -x/y, and I was curious as to how that would be done as well, if you don't mind.

Answer 15

isn't the derivative of y, y'? so would you plug in y'?

Answer 16

OK, we'll see both the methods
\[2+2(y')^2+2yy''=0\]
Now solve for y'' from here

Answer 17

@lizlozada ??

Answer 18

sorry i was helping my little sister with her essay

Answer 19

no problem. Take you time

Answer 20

isn't it -x/y because you factor a y' and divide by 2y?

Answer 21

but the first term is 2. It doesn't have a y'

Answer 22

oh right i didn't notice that. take way two from both sides then factor out a y'?

Answer 23

Solve for y'' from here
\[2+2(y')^2+2yy''=0\]

Answer 24

so you don't take away 2 from both sides?

Answer 25

Take 2 away, take y'' on one side

Answer 26

then you factor out a y' from the left?

Answer 27

\[1+(y')^2+y' y"=0\]
\[y' y''=-(1+y')^2
\]
divide both sides by y'

Answer 28

Sorry
this is the correct expression
\[y' y''=-(1+y'^2)\]

Answer 29

why is it 1?

Answer 30

We have
\[2+2(y')^2+2y' y''=0\]
divide both sides by 2
\[1+(y')^2+y' y"=0\]

Answer 31

oh ok. what's the difference between y' & y"?

Answer 32

\[y'=\frac {dy}{dx}\]
\[y''=\frac {d^2y}{dx^2}\]

Answer 33

@lizlozada I'll be back in 10 minutes. You try this until then

Answer 34

ok

Answer 35

ok i am very confused. would you mind teaching me how to take the derivative of -x/y? i feel that's simpler,

Answer 36

Cool. Let's work that way

Answer 37

thank you!

Answer 38

we have
\[\frac {dy}{dx}=\frac{-x}{y}\]
Let's take derivative both sides, we get
\[\frac{d^2y}{dx^2}=\frac{d}{dx} (\frac {-x}{y})\]
do you know division rule of differentiation ?

Answer 39

loDhi-hiDlo/ lolo?

Answer 40

yes, apply that here

Answer 41

y(1)-x(y')/y^2

Answer 42

good, so we have
\[\frac{d^2y}{dx^2}=\frac{y-xy'}{y^2}\]
Let's find value of this at x=4,
from x^2+y^2=25
what's value of y when x=4?

Answer 43

which equation do i plug it into?

Answer 44

x^2+y^2=25
this is the one which relates x and y

Answer 45

then why do we get d^2y/dx^2 if we just put 4 in the og equation?

Answer 46

we need to find the value of \(\frac{d^2y}{dx^2}\) at x=4, not value of y

Answer 47

16+y^2=25
y^2=9
y=3

Answer 48

then?

Answer 49

Now find value of y' at x=4 and y=3

Answer 50

\[\frac {dy}{dx}=\frac{-x}{y}\]

Answer 51

-4/3

Answer 52

Good :)
Now substitute these values here
to find d^2y/dx^2
\[\frac{d^2y}{dx^2}=\frac{y-xy'}{y^2}\]

Answer 53

3-(4)(-4/3)/9
(4/3)/9
4/27?

Answer 54

Check your calculations again :)

Answer 55

ok sorry.(3+16/3)/9
=1

Answer 56

\[\large\frac{3+\frac{16}{3}}9\]
\[\large\frac{\frac{9+16}{3}}9\]
\[\large\frac{\frac{25}{3}}9\]
\[\large\frac{25}{27}\]

Answer 57

@lizlozada read the whole post and do let me know if you have a doubt

Answer 58

yes sorry I was rushing my math. everything is crystal clear, thank you so much.

Answer 59

Are you certain?

Answer 60

yes, i was thinking 16+9 is 27. so it would be 27/3=9, 9/9=1.
i fully understand, thank you.

Answer 61

Welcome :D