Question

The point P (pi/2 , 0) lies on the curve y = xcosx .

Find the equation of the normal to the curve at the point P.

Answer 1

You can't assume the line is perfectly perpendicular to the curve so we assume the following. Assume another line is TANGENT to the curve and passes through P(pi/2, 0). Find the slope of that tangent line by taking the derivative of y=xcosx, then substitute pi/2 for x in the derivative, the new value gotten is the SLOPE of the tangent line. The normal is perfectly perpendicular to the tangent line and we know that the product of the slopes of perpendicualr lines is -1 so use m1m2=-1 to find the slope of the normal line m1 is the slope of the tangent, so solve for m2.

Answer 2

First of all, how do I differentiate xcosx?

Answer 3

We use the multiplication rule
\[ \frac{d}{dx}(x*Cosx)=1*Cos(x)+x*(-Sinx)= Cosx-x*Sinx\]

Answer 4

So when P = (pi/2), dy/dx = cos pi/2 - pi/2 sin pi/2

Answer 5

= -1.57?

Answer 6

Yup

Answer 7

Is it alright to leave it as -1.57 or does it have to be -pi/2?

Answer 8

Well its easier to leave it as -pi/2
Cos(pi/2)=0 and sin(pi/2)=1
So \[ \cos( \frac{\pi}{2})-\frac{\pi}{2} \sin(\frac{\pi}{2})= 0-\frac{\pi}{2}(1)=-\frac{\pi}{2}\]

Answer 9

So obviously the slope of the normal is \(\large \frac{ 2}{\pi}\) since
\[ \frac{ 2}{\pi}*-\frac{ \pi}{2}=-1\]

Answer 10

Why is it 2/pi again? I forgot

Answer 11

pi/2*m2=-1, what is m2 @Lii

Answer 12

I don't know lol

Answer 13

-0.63?

Answer 14

If we have the equation. \[-\frac{ \pi }{ 2 } \times m_{2} =-1\] How can you solve for m2?

Answer 15

Make m2 the subject

Answer 16

so 0.63 so therefore 2/pi

Answer 17

Exactly! that answers your question.

Answer 18

Ok so y-0 = 2/pi (x-pi/2)

Answer 19

y = 2/pi x - 1

Answer 20

Yes.

Answer 21

Thanks but as -pi/2 = -1.57, would substituting -1.57 still get the right answer?

Answer 22

@Isaiah.Feynman so if I substituted -1.57 instead:

y-0 = -1.57(x-pi/2)
y-0 = -1.57x + 2.46
y = -1.57x + 2.46, is that right or wrong?

Answer 23

I think it would be wrong.

Answer 24

Oh ok, well thanks @Isaiah.Feynman and @BlackLabel