# Symmetric equations of tangent lines to curves

1. Aug 2, 2008

### 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 ;)

2. Aug 3, 2008

### futurebird

$$z=x^{2} + y^{2}$$ is a http://mathworld.wolfram.com/Paraboloid.html" [Broken]. $$x+y+6z=33$$ is a plane. Their intersection must be a point, no intersection, or a curve. (1,2,5) satisfies both equations so it is the point of intersection, or it is on the curve formed by the intersection.

Last edited by a moderator: May 3, 2017
3. Aug 3, 2008

### CalleighMay

k so how do i find "symmetric" equations of the "tangent line to the curve of intersection of the surfaces" (BIG WORDS) at the given point?? I'm confused as to what exactly it wants me to do...

4. Aug 4, 2008

### CalleighMay

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

5. Aug 4, 2008

### carlodelmundo

Calleigh,

Just a quick tip with the Larson--"Calculus 8th Edition" Textbook. If you don't know already, a great resource that accompanies the book is http://www.calcchat.com/ . It has answers to the ODD number questions (with step-by-step solutions...not the final product you'll see in the back of the book). If you subscribe to their service, a "Assistant" (prob. a grad TA or something) will answer your questions on ALL of the ODD problems (so in effect, it's good for practice in order to tackle the problem).

The assistant is actually a human being.... however, their summer hours are only from 1PM-4PM EST.

I am not advertisting. Just stating a resource--I'm using their service recently as I'm only in Chapter 7.1.

Good luck

6. Aug 4, 2008

### CalleighMay

thanks for that link carlodelmundo! ;)

Is there a way to find worked out soln's to the even ones? Darn that's prolly why my professor only gives us even ones to do since all the odd ones are online lol