Symmetric equations of tangent lines to curves

[CalleighMay]

Hello, my name is Calleigh and i am new to the forum! I am in Calculus II and have a few questions on some problems. I am using the textbook Calculus 8th edition by Larson, Hostetler and Edwards. Could someone please help me?

 

The problem is on pg 950 in chapter 13.7 in the text, number 46. It reads:

 

a. Find symmetric equations of the tangent line to the curve of intersection of the surfaces at the given point.

b. Find the cosine of the angle between the gradient vectors at this point.

c. State whether or not the surfaces are orthogonal at the point of intersection.

 

And they give:

z=x^2+y^2, and x+y+6z=33 and the pt (1,2,5).

 

 

My first problem is understanding how to draw this with the z thing. What's the tangent line to this curve (wait, what curve??) and what does it means when it asks for "the cosine of the angle". And how do i tell if they're orthogonal, do i use the dob (sp?) product or something like that? I'm pretty lost as you can tell.

 

Any help would be greatly appreciated! Thanks guyssss

[Dave]

I don't have the time to fully explain the problem, but essentially you have two surfaces here which intersect one another at a number of points. The idea here is to define a curve by the intersection of the two surfaces.

 

Drawing in 3D is pretty tricky. The easiest thing to do is to set particular values for x and y, and then figure out what z does. So, for example, with the first surface, if I set x=0, I get z = y^2 which is a simple parabola. Similarly, set y=0 and you get a parabola. In fact, that surface is known as a paraboloid; the rotation of a parabola around the z-axis. The second surface is clearly a plane.

[CalleighMay]

Thanks for the reply Dave! ;P

 

So ummm how do i calculate to find what i'm looking for? And what's the "cosine of the angle"? (what angle...?)

 

well my friends and i were going over these problems tonight and this is what we have so far...

 

F(x,y,z)=x^2+y^2-z

F^(x,y,z)=2xi+2yj-k

(1,2,5)=2i+4j-k

G(x,y,z)=x+y+6z-33

G^(x,y,z)=i+j+6k

 

The cross product of these two gradients which is a vector tangent to both surfaces at the point (1,2,5). We did the cross product:

 

^F x ^G= 25i-13j-2k

 

direction numbers for part a are 25, -13, -2

 

symmetric equations:

(x-1)/25, (y-2)/-13, (z-5)/-2

 

and cos(theta)= (I ^F x ^G I) / (II ^F II x II ^G II) which equals 0 so it's orthogonal.

 

 

How's that look? =D