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

 

[tkhunny]

You can be expected to know that the dot product of two normalized vectors produces the cosine of the angle between the two vectors.

 

[Deleted member 4993]

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?? Two surfaces intersect along a curve - like two curves intersect at a point

 

[CalleighMay]

so what do i do? lol

 

[stapel]

so what do i do? lol

Do you know what "vectors" are, or what "dot" and "cross" products are? Do you know what it means for two things (say, lines) to be "perpendicular" or "orthogonal" to each other?

Eliz.

 

[CalleighMay]

Thanks for the response Eliz!

Oh yeah, i know vectors are basically lines with directions, ie 2i+4j+5k means 2 in the x dir, 4 in the y dir and 5 in the z dir. I used vectors a lot in physics, just in calc it seems to confuse me.

We sort of "covered" aka i've heard of, the dot product, but i have no idea how to use it or what it does.

And i know that perpendicular is when lines intersect at 90 degrees and isn't that the same as orthogonal?

My professor has given us some problems to attempt to do even though it's 3 chapters ahead. We will be covering them next semester, he just wants to give us a taste of what we're in for. He's not grading these problems it's sort of extra credit. I just want to be able to impress him! Thanks for any help!

 

[stapel]

We sort of "covered" aka i've heard of, the dot product, but i have no idea how to use it or what it does....

A good first step in attempting to teach yourself new material will be to look up the terms used, and study what they mean and how they're used "in context". For instance:

. . . . .Google results for "dot cross product"

As the tutors' replies have indicated, you need to learn all this material before you can attempt this particular exercise: the exercise assumes that you already have a good grasp of this material. You'll also want to study some online lessons regarding "symmetric equations", intersections of surfaces, and gradient vectors, along with whatever other terms you encounter which have not yet been taught in class. :idea:

Good luck!

Eliz.

 

[galactus]

\(\displaystyle F(x,y,z)=x^{2}+y^{2}-z\)

\(\displaystyle {\nabla}F(x,y,z)=2xi+2yj-k\)

\(\displaystyle {\nabla}(1,2,5)=2i+4j-k\)

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

\(\displaystyle {\nabla}G(x,y,z)=i+j+6k\)

\(\displaystyle {\nabla}(1,2,5)=i+j+6k\)

Now, the cross product of these two gradients is a vector that is tangent to both surfaces at the point (1,2,5).

So, take the cross product:

part a: \(\displaystyle {\nabla}F \times {\nabla}G=\begin{vmatrix}i&j&k\\2&4&-1\\1&1&6\end{vmatrix}=25i-13j-2k\)

Direction numbers:

\(\displaystyle 25, \;\ -13, \;\ -2\)

Symmetric equations:

\(\displaystyle \frac{x-1}{25}, \;\ \frac{y-2}{-13}, \;\ \frac{z-5}{-2}\)

part b:

\(\displaystyle cos{\theta}=\frac{|{\nabla}F\cdot {\nabla}G|}{||{\nabla}F||||{\nabla}G||}=0\)

Therefore, it is orthogonal.

Now, there is a complete example for future problems like this. Look in your book for a worked example.

Larson and Hostetler make just about one of the best calc books.

 

[CalleighMay]

So that's the answer!? =D

Yeah this book is one of the better ones that i've had throughout my education for math, it has a few examples that are helpful yet skip some steps. Nothing like one-on-one help! =D Thanks!