Parametric curves at right angles.

[mesty]

Hi all,

I have a homework problem which I am struggling with.

The question is as follows:

Consider the two curves C1 and C2 given parametrically.
C1 : x(t)=1−t, y(t)=1−(1/t), z(t)=2t^2 −2
C2 : x(x)=x, y(x)=sinx, z(x)=2cosx−2

Show that C1and C2 intersect at right angles at the origin.

Do I just let t and x equal zero to get two vectors and show the two vectors are at right angles?

If I let t equal zero then y(t)=1-(1/0), what happens with the divide by zero?

Am I on the right track? If not how do I solve this question.

Thanks!

 

[Deleted member 4993]

Hi all,

I have a homework problem which I am struggling with.

The question is as follows:

Consider the two curves C1 and C2 given parametrically.
C1 : x(t)=1−t, y(t)=1−(1/t), z(t)=2t^2 −2
C2 : x(x)=x, y(x)=sinx, z(x)=2cosx−2

Show that C1and C2 intersect at right angles at the origin.

Do I just let t and x equal zero to get two vectors and show the two vectors are at right angles?

If I let t equal zero then y(t)=1-(1/0), what happens with the divide by zero?

Am I on the right track? If not how do I solve this question.

Thanks!

\(\displaystyle t \ne 0\) at the origin!

at the origin you have the condition x = y = z = 0 → at what value of 't' does that happen?

 

[mesty]

When t=1. Does that mean we have the vector (1-1,1/(1/1),2-2)=(0,0,0) and then we just need to do the same of the other curve and then prove that the two vectors are perpendicular?

Am I on the right track?

Thanks

 

[Deleted member 4993]

When t=1. Does that mean we have the vector (1-1,1/(1/1),2-2)=(0,0,0) and then we just need to do the same of the other curve and then prove that the two vectors are perpendicular?

Am I on the right track?

Thanks

(0,0,0) is not a vector - it is the co-ordinate of a given point on the curve C₁.

It shows that the origin (0,0,0) is a point located on the curve.

Now you need to show that (0,0,0) is a point on the curve C₂.

Then you need to calculate the gradients of the curves at the origin - and show those to be perpendicular to each other.

 

[pka]

The question is as follows:
Consider the two curves C1 and C2 given parametrically.
C1 : x(t)=1−t, y(t)=1−(1/t), z(t)=2t^2 −2
C2 : x(x)=x, y(x)=sinx, z(x)=2cosx−2
Show that C1and C2 intersect at right angles at the origin.

My view is that this is a poorly written question.
Using two different parameters \(\displaystyle c_1(t)=<1-t,1-(1/t),2t^2-2>~\&~c_2(s)=<s,\sin(s),2\cos(s)-2>\)

Now \(\displaystyle c_1(1)=(0,0,0)=c_2(0)\), so the curves intersect at \(\displaystyle (0,0,0)\).

The derivatives, \(\displaystyle (c_1)'(t)=<-1,(1/t^2),4t>~\&~(c_2)'(s)=<1,\cos(s),-2\sin(s)>\) are the directions.

So \(\displaystyle (c_1)'(1)=<-1,1,4>~\&~(c_2)'(0)=<1,1,0>\). Are they perpendicular?

 

[mesty]

Now \(\displaystyle c_1(1)=(0,0,0)=c_2(0)\), so the curves intersect at \(\displaystyle (0,0,0)\).

I dont understand how you can say this? Could you elaborate.

 

[HallsofIvy]

You say that the parametric equations are x(t)=1−t, y(t)=1−(1/t), z(t)=2t^2 −2
and x(s)= s, y(s)= sin(s), z(s)=2cos(x)−2. They will intersect, of course, where the x, y, and z coordinates are the same: x= 1- t= s, y= 1- (1/t)= sin(x), and \(\displaystyle z= 2t^2- 2= 2cos(t)- 2\). Since s= 1- t, the second equation is the same as 1- (1/t)= sin(1- t). There are, in fact, a number of solutions to this but the easiest one is t= 1. Further, that satisfies the third equation which, with s= 1- t is \(\displaystyle 2t^1- 2= 2cos(0)- 2\) while other values do not. From that, x= 1-1= 0, y= 1- (1/1)= 0, z= 2(1)- 2= 0.