Is My Solution for 3D Triangle Area Correct?

[DODGEVIPER13]

Now I get y^2-y

 

[gneill]

Dy=2x

$y = x^2$
$dy = 2x dx$

 

[gneill]

Now I get y^2-y

Show how. Your final result should be numeric.

 

[DODGEVIPER13]

What I did is take the integral from 1 to 0 of y dx and found it to be -y and then evaluated the integral from 0 to 1
Of 2xy^2 dx and got y^2 should I have turned y into x^2 and y^2 into x^4 hmm if I do that I get 0

 

[gneill]

What I did is take the integral from 1 to 0 of y dx and found it to be -y and then evaluated the integral from 0 to 1
Of 2xy^2 dx and got y^2 should I have turned y into x^2 and y^2 into x^4 hmm if I do that I get 0

I don't know why you'd do that first integral. The idea is to cast the dot product into terms of a single variable, and x is convenient. To do that you must replace the $y^2$ of the H function using $y=x^2$.

Now, the path you want L to follow is along the other function $y=x^2$. So then in terms of x, $dy = 2x dx$. Make the substitutions and carry out the dot product. Then integrate over x. By substituting y and dy, you need only integrate over the one variable, x.

 

[DODGEVIPER13]

Integrate from 1 to 0 of x^2 dx and I get -1/3 evaluate the second integral from 0 to 1 of 2x^5 dx and get 1/3 added together I get 0 which is not right?

 

[DODGEVIPER13]

Or is it just one integral once I do the substitution from 1 to 0 of x^2+2x^5 which would be -2/3

 

[gneill]

Or is it just one integral once I do the substitution from 1 to 0 of x^2+2x^5 which would be -2/3

Once you've made the substitutions and performed the dot product (collecting the terms into a single sum) then there is only one integral to perform over x from 0 to 1.

 

[DODGEVIPER13]

So 2/3

 

[DODGEVIPER13]

Integration done would be x^3/3+x^6/3 which when evaluated from 0 to 1 I get 2/3

 

[gneill]

Yup, that's what I get.

 

[DODGEVIPER13]

Ok man so do you suggest I ask separate questions for the rest also I will give a thanks? I was wondering if I add one thanks does it count for all the posts.

 

[gneill]

Ok man so do you suggest I ask separate questions for the rest also I will give a thanks? I was wondering if I add one thanks does it count for all the posts.

One question per thread is the preferred format; it avoids the confusion that can ensue if multiple people start addressing different different questions in random order.

Presumably one "Thanks" covers whatever you want... there are no guidelines for Thanks that I'm aware of, and they don't appear to alter the status of a post or thread in any way. They're just a friendly tip-of-the-hat to signify appreciation for help given.

Cheers.

 

[DODGEVIPER13]

Hey I know this is a really old question but I never got the chance to check my angle answer I got 83.62 what I did was cos(theta)= (A dot B)/(abs(a)Abs(B)) then cos^-1(1/9)=83.62 degrees

 

[DODGEVIPER13]

Since its really old just so everyone knows I am referring to probem 1.3 on the pdf that I have uploaded

 

[DODGEVIPER13]

well nvm that does not sum to 180 degrees but I think I may have solved it I got 76.86,57.31, and 45.81 is that correct?

 

[DODGEVIPER13]

sorry one final check on 1.2 the perpendicular part, I already have the parallel part. I found alpha= 4 and beta=1 is this right?

 

[gneill]

well nvm that does not sum to 180 degrees but I think I may have solved it I got 76.86,57.31, and 45.81 is that correct?

Yes, those angles look okay.

 

[gneill]

sorry one final check on 1.2 the perpendicular part, I already have the parallel part. I found alpha= 4 and beta=1 is this right?

You should be able to test your results. What's the test for perpendicularity for vectors?

 

[DODGEVIPER13]

If their dot product equals zero then they are perpendicular which my numbers seem to pass. As long as I'm doing the dot product right which I am pretty sure I am multiplying like values together and adding them.

 

[gneill]

If their dot product equals zero then they are perpendicular which my numbers seem to pass. As long as I'm doing the dot product right which I am pretty sure I am multiplying like values together and adding them.

Can you show that work?

 

[DODGEVIPER13]

A=AlphaUx+3Uy-2Uz and B=4Ux+betaUy+8Uz then by observation I noticed if I plugged 1 in for alpha then when I multiplied 4Ux and 1 Ux I would get 4. I then saw that is I plugged 4 in for beta then when 3Uy and 4Uy multiplied I would get 12. These added together gave me 16 and when -2Uz and 8Uz I get -16 when added together I get 0 which I assumed was right but I'm guessing its not.

 

[gneill]

A=AlphaUx+3Uy-2Uz and B=4Ux+betaUy+8Uz then by observation I noticed if I plugged 1 in for alpha then when I multiplied 4Ux and 1 Ux I would get 4. I then saw that is I plugged 4 in for beta then when 3Uy and 4Uy multiplied I would get 12. These added together gave me 16 and when -2Uz and 8Uz I get -16 when added together I get 0 which I assumed was right but I'm guessing its not.

Okay, previously you claimed $\alpha = 4$ and $\beta = 1$. Now you've swapped the values after checking. Always check!

While this (new) pair of values is one particular solution, you can get a general solution by expanding the [STRIKE]cross[/STRIKE] dot product symbolically and setting it to zero. You'll notice it yields one equation in two unknowns. What does that tell you?

Edit: Oops, I meant dot product, of course!

 

[DODGEVIPER13]

Ok I got something wonky trying to do the cross product of that I can't upload it until around 1 so ill do it then but typically when you have 2 unknowns I would need a second equation correct? They must then both multiply times the constants to form the 0 maybe?

 

[gneill]

Ok I got something wonky trying to do the cross product of that I can't upload it until around 1 so ill do it then but typically when you have 2 unknowns I would need a second equation correct? They must then both multiply times the constants to form the 0 maybe?

Check my edit of my previous post: I meant DOT product, not CROSS product, of course.

If a system of equations is underdetermined (more unknowns than equations), then you can have an infinite number of solutions but a fixed relationship between the variables...

 

[DODGEVIPER13]

ah makes sense so this obviously isn't the only solution but nonetheless one that will be correct right?

 

[gneill]

ah makes sense so this obviously isn't the only solution but nonetheless one that will be correct right?

Well, certainly not wrong, but at best incomplete