Calc 3 question - Tilted Forum Project Discussion Community

krwlz · 10-13-2004, 07:39 AM

This one was on a test I just took, and I had to leave it blank, its driving me insane:

Find the point of intersection between a line with the equations:
X/2 = (Y-9) = (Z+1)/2

and the Plane:
x-y-z=4

Now, I dont necesarilly need the answer, but how the fuck do I solve this? Is it simply guess and check? I tried using the parametirc form of the line equations to set them equal to the the plane, but god damn, Im confused.

Also, the other one I had a problem with (since I havnt done integration or derivation all summer, and it was on the test)

take the derivative of t(cos(t))

Phage · 10-13-2004, 08:00 AM

Now I have not taken Calc 3, but I am guessing from what we are doing in my multivariable calc class...
It is really easy to find the intersection of two lines right? So, to find the intersection of a line with a plane might it be easier to find the equation for (three?) lines on the plane, one on each dimension, and find where they intersect then combine those results?

Ydde · 10-13-2004, 08:02 AM

well if u look at it, you have to solve for 3 variables. X, Y, Z
You got enough equations to solve for the variables.

Express Y in terms of X only and Z in terms of X only.
Y = X/2 + 9 and Z = 2(Y-9) - 1

Now subsitute them into the plane equations. Now you have only one variable to solve, that is X.

than you can sovle for Y and Z !

hope that helps

AngelicVampire · 10-13-2004, 08:26 AM

Firstly you can solve purely for X (or Y or Z) by substituting the Y and Z variables for some function of X...

X - (X+18)/2 - (X-1) = 4
=> -X-18 = 6
=> X = -24

then X/2 = Y-9, Y = -3

And Z = X-1 = -25

Hopefully the maths is right but the idea should be.

mathmo · 10-13-2004, 01:24 PM

No-one seems to have answered the last question, the derivative of t*cos(t)

This just comes from the product rule. That is:

d/dt (f(t)*g(t)) = f'(t)*g(t) + f(t)*g'(t)

where f and g are functions in t, in this case let f(t) = t and g(t) = cos(t)
and f'(t) denotes the derivative of f(t), similarly for g(t). Hence in our case, f'(t) = d/dt (t) = 1, g'(t) = d/dt (cos(t)) = -sin(t)

So, d/dt t*cos(t) = cos(t) + t*(-sin(t)) = cos(t) - t*sin(t)

Sorry for trying to use so much math notation, so I hope it is clear.

krwlz · 10-13-2004, 02:48 PM

Yea, it is, i remember that now, I just havnt used that shit in a while. As to the Plane and line, that all sounds right, and I thought thats what I was trying to do, but you know how it goes when you start to get frustrated with something.

THing is... In a test, you cant walk back in a half hour later, and finish up that last problem.

10-13-2004, 07:39 AM	#1 (permalink)
krwlz Fledgling Dead Head Location: Clarkson U.	Calc 3 question This one was on a test I just took, and I had to leave it blank, its driving me insane: Find the point of intersection between a line with the equations: X/2 = (Y-9) = (Z+1)/2 and the Plane: x-y-z=4 Now, I dont necesarilly need the answer, but how the fuck do I solve this? Is it simply guess and check? I tried using the parametirc form of the line equations to set them equal to the the plane, but god damn, Im confused. Also, the other one I had a problem with (since I havnt done integration or derivation all summer, and it was on the test) take the derivative of t(cos(t))

10-13-2004, 08:00 AM	#2 (permalink)
Phage Insane	Now I have not taken Calc 3, but I am guessing from what we are doing in my multivariable calc class... It is really easy to find the intersection of two lines right? So, to find the intersection of a line with a plane might it be easier to find the equation for (three?) lines on the plane, one on each dimension, and find where they intersect then combine those results?

10-13-2004, 08:02 AM	#3 (permalink)
Ydde Tilted Location: Singapore	well if u look at it, you have to solve for 3 variables. X, Y, Z You got enough equations to solve for the variables. Express Y in terms of X only and Z in terms of X only. Y = X/2 + 9 and Z = 2(Y-9) - 1 Now subsitute them into the plane equations. Now you have only one variable to solve, that is X. than you can sovle for Y and Z ! hope that helps

10-13-2004, 08:26 AM	#4 (permalink)
AngelicVampire Insane	Firstly you can solve purely for X (or Y or Z) by substituting the Y and Z variables for some function of X... X - (X+18)/2 - (X-1) = 4 => -X-18 = 6 => X = -24 then X/2 = Y-9, Y = -3 And Z = X-1 = -25 Hopefully the maths is right but the idea should be.

10-13-2004, 01:24 PM	#5 (permalink)
mathmo Tilted Location: London, UK	No-one seems to have answered the last question, the derivative of tcos(t) This just comes from the product rule. That is: d/dt (f(t)g(t)) = f'(t)g(t) + f(t)g'(t) where f and g are functions in t, in this case let f(t) = t and g(t) = cos(t) and f'(t) denotes the derivative of f(t), similarly for g(t). Hence in our case, f'(t) = d/dt (t) = 1, g'(t) = d/dt (cos(t)) = -sin(t) So, d/dt tcos(t) = cos(t) + t(-sin(t)) = cos(t) - t*sin(t) Sorry for trying to use so much math notation, so I hope it is clear.

10-13-2004, 02:48 PM	#6 (permalink)
krwlz Fledgling Dead Head Location: Clarkson U.	Yea, it is, i remember that now, I just havnt used that shit in a while. As to the Plane and line, that all sounds right, and I thought thats what I was trying to do, but you know how it goes when you start to get frustrated with something. THing is... In a test, you cant walk back in a half hour later, and finish up that last problem.