Finding the Normal and Intersection Points on a Parametric Circle Curve

[thereddevils]

Homework Statement

Find the equation of normal to curve x=cos b , y=sin b at the point where b=pi/4 . Find the coordinates of the point on the curve where the normal meets the curve again .

Homework Equations

The Attempt at a Solution

Gradient of normal :

dx/db=-sin b

dy/db=cos b

dy/dx=-cot b

gradient of normal = tan b , when b=pi/4 , gradient of normal =1

The normal cuts the curve at point (1/sqrt(2) , 1/sqrt(2))

equation of normal is y-1/sqrt(2)=1(x-1/sqrt(2))

y=x --- 1

The cartesian equation of the curve is cos^(-1) x=sin^(-1) y

sin y=cos x --- 2

combine 1 and 2 , sin x=cos x

tan x=0

x=0 , pi , 2pi

Since the domain is not mentioned , i take it as 0<=x<=2pi

Therefore , the normal would cut the curve at points (0,pi/2) , (pi,3pi/2), (2pi, 5pi/2)

Am i correct ?

 

[m.w.lever]

I am confused by eq. 1 and 2. You converted (1) from cartesian to "wonky" cartesian. You have the equation of the normal line as y=x, which is correct, but that second equation seems to come out of nowhere.

What does the original function actually represent? If you graph the curve, and y=x (the normal line) the answers become more clear.

As a side note, limiting the domain might be a bad idea ;-)

 

[thereddevils]

I am confused by eq. 1 and 2. You converted (1) from cartesian to "wonky" cartesian. You have the equation of the normal line as y=x, which is correct, but that second equation seems to come out of nowhere.

What does the original function actually represent? If you graph the curve, and y=x (the normal line) the answers become more clear.

As a side note, limiting the domain might be a bad idea ;-)

ok , so how do i get the equation of the curve given its parametric equation ? I tried to eliminate the parameters and got that second equation .

 

[m.w.lever]

Well, the curve given is a circle with radius 1.
This can be verified: The equation of a circle is
x²+y²=r²
since you were given:
x=cos(b)
y=sin(b)
x²+y²=r² goes to:
cos²(b)+sin²(b)=r²
1=r²
1=r

If you draw the circle and the line y=x, which according to the problem is:
sin(b)=cos(b)
sin(b)/cos(b)=1
tan(b)=1

As for the intersections:
tan(b)=1=x²+y² I think you had zero here by mistake.
If you check your unit circle, this occurs at Pi/4 +kPi, where k is an integer. (This works for all real domains, not just zero to 2Pi)

So you were on the right path from the get go, but the extra transform threw things off.

 

[thereddevils]

Well, the curve given is a circle with radius 1.
This can be verified: The equation of a circle is
x²+y²=r²
since you were given:
x=cos(b)
y=sin(b)
x²+y²=r² goes to:
cos²(b)+sin²(b)=r²
1=r²
1=r

If you draw the circle and the line y=x, which according to the problem is:
sin(b)=cos(b)
sin(b)/cos(b)=1
tan(b)=1

As for the intersections:
tan(b)=1=x²+y² I think you had zero here by mistake.
If you check your unit circle, this occurs at Pi/4 +kPi, where k is an integer. (This works for all real domains, not just zero to 2Pi)

So you were on the right path from the get go, but the extra transform threw things off.

thanks a tons , m.w.Clever

 

[m.w.lever]

You're welcome, but no C in lever :) It can be tricky to visualize a lot of parametric problems, but in some cases (like this one) it really helps!

 

[Mark44]

thereddevils,
You really should post calculus problems in the Calculus & Beyond section, not the Precalculus section

 

[thereddevils]

sorry to trouble you with a few more questions ,

(1) b= pi/4+kpi , is this the coordinate ? Or rather the coordinate is x=cos pi/4+kpi and
y= sin pi/4+kpi ?

because from y=x , x^2+y^2=1 , i could have solved for x and y but that's for b=pi/4 only right ?

Mark , sorry for posting on the wrong subforum coz i thought this is just a high school question .

 

[Mark44]

Doesn't matter if it's high school or college or whatever. Calculus questions should go to the Calculus & Beyond section.

The line y = x comes about because the problem asks for the normal on the parametric curve at b = pi/4. This point corresponds to the point (sqrt(2)/2, sqrt(2)/2) on the circle. Because the curve is a circle, the normal to the curve intersects the circle at a point directly across the circle, at (-sqrt(2)/2, -sqrt(2)/2). The x and y values in these points are the coordinates.