Question

y = cos(x^2), dy/dx = -2xsin(x^2)

FInd the co-ordinates of M such that the x-coordinate is between 0 and 1, and that the normal at that point goes through the origin.

Answer 1

This is what i've done till now: I've found that the equation of the tangent is y = mx +c and the equation of the normal is y = -x/m

Answer 2

then I put in the dy/dw as the m

Answer 3

and put in y as cos(x^2)

Answer 4

But after simplifying, i'm getting (for the normal equation because that doesn't have a constant) cos(x^2)*sin(x^2) = 1/2

Answer 5

And now I have no idea how to proceed :P Commence discussion.......NOW.

Answer 6

\[y = cosx^2, \frac{ dy }{ dx } = -2xsinx^2, cosx^2sinx^2 = \frac{ 1 }{ 2 }\]

Answer 7

@terenzreignz

Answer 8

Okay, so, as you said, the equation of the normal line is
\[\huge y = -\frac{x}m\]right?

Answer 9

yup.

Answer 10

And then, substituting dy/dx for m...
\[\huge y = \frac1{2\sin(x^2)}\]

Answer 11

yes, but then I also substituted y as cos(x^2)

Answer 12

hang on.
The equation of the normal line is
\[\huge y = -\frac{x}m + c\]for some constant c

Answer 13

nope, it passes through the origin.

Answer 14

Okay, say we have a differentiable function y = f(x)
At any point \(\large (x_o , f(x_0))\) the equation of the normal line is
\[\Large y = \frac{x_o-x}{f'(x_o)}+f(x_o)\] right?

Answer 15

yeahh, I think so (sorry all the letter are confusing me but I get it :P)

Answer 16

Now, since it has to pass through the origin, we let x = y = 0
\[\Large 0 = \frac{x_o}{f'(x_o)}+f(x_o)\]
For convenience, set
\[\large x_o = c\]
\[\Large 0 = -\frac{1}{2\sin(c^2)}+\cos(c^2)\]
And now, it's just a matter of solving for c... hopefully.

Answer 17

Well, let's see...
\[\large \frac1{2\sin(c^2)}=\cos(c^2)\\\large 1 = 2\sin(c^2)\cos(c^2)\\\large1=\sin(2c^2)\\ \\ \\ \large 2c^2=\frac{\pi}2 \\ \large c^2 = \frac{\pi}4 \\ \\ \huge \color{blue}{c=\frac{\sqrt \pi}{2}}\]


How about that, @tanvidais13 ?
Not bad, huh? ;)

Answer 18

So the point is going to be (c , f(c) )

\[\huge \left(\frac{\sqrt \pi}{2}, \cos\left(\frac{\sqrt{\pi^2}}{2^2}\right) \right)\]
\[\huge \left(\frac{\sqrt \pi}{2}, \cos\left(\frac{\pi}4\right) \right)\]

Now, do the honours, and do the final step :D

Answer 19

So basically you ended up in pretty much the same place I did, looking a lot fancier :P
OMG I THOUGH WE COULDN'T USE THE SIN RULE BECAUSE IT WAS A SQUARE
Why am I such an idiot??

Answer 20

I am no longer disoriented. I credit the taco I just ate :D
cheers!

Answer 21

Ohhh, I wish I could give you more than one medal. You, sir, are a life saver. Really.