#$&*
course mth 277
12/7 11
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
query_10_1
*********************************************
Question: Find the domain of F(t) X G(t) when F(t) = t^2 i - (t+2)j + (t-1)k and G(t) = (1/(t+2))i + (t-5)j + sqrt(t) k.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution: After I do the cross product I am unsure what to do should I just set the new equation equal to zero and solve for t? There is not a good example of this in the book and there is not a qa for 10.1.
@&
G(t) is not defined when t < 0.
It's also not defined when t = -2, but that's already covered by the restriction that t can't be < 0.
*@
???????????????????????????????????????????????????????
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question: Describe the graph of G(t) = (sin t)i + (cos t)j + (4/3)k
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
The graph would be a line that lies on the surface of circular cylinder. It is a circle because of the (sint)i+(cost)j and it forms a cylinder because it has a slope of (4/3) in the positive k direction.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):Okay
------------------------------------------------
Self-critique rating:3
@&
Good.
*@
*********************************************
Question: Given F(t)= (t)i - 5(e^t)j +(t^3)k, G(t) = ti - (1/t)k and H(t) = (t*sin t)i + (e^-t)j, find H(t) dot [G(t) X F(t)]
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution: Im okay with dot and cross products.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):okay
------------------------------------------------
Self-critique rating:3
*********************************************
Question: Find a vector function F whose graph is the curve given by the equation x/5 = (y-3)/6 = (z+2)/4.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution: I just set x/5=0, (y-3)/6=0, (z+2)/4= 0 so that the vector function F would equal 3j-2z.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):okay
@&
You want a vector function `F(t) = x(t) `i + y(t) `j + z(t) `k.
If you let t = x / 5, then x = 5 t so your x(t) function could just be x(t) = 5 t.
Then your equation becomes
t = (y - 3) / 6 = (z + 2) / 4. sp
(y-3) / 6 = t
and
(z + 2) / 4 = t.
Thus
y = 6 t + 3
and
z = 4 t - 2,
so your function would be
F(t) = t `i + (6t + 3) `j + (4 t - 2) `k.
*@
------------------------------------------------
Self-critique rating:3
*********************************************
Question: Find the limit as t -> 2 of ((t^4-2)/(t-2))i + ((t^2-4)/(t^2-2t))j + ((t^2 + 3)e^(t-2))k.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution: you would take the limit as t approaches 2 of the i,j,k components.
limF(t) as t approaches 2 would approach zero according to l’hospitals rule because the denominator is undefined.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):im still a little unsure.
------------------------------------------------
Self-critique rating:3
@&
By l'Hopital's rule the limiting value of (t^4 - 2) / (t - 2), as t -> 2, is the same as the limiting value of 3 t^3 / 1, obtained from the derivatives of the numerator and denominator. So the limit would be 3 * 2^4 / 1 = 48.
(t^2 - 4) / (t^2 - 2 t) factors into
(t - 2) (t + 2) / (t ( t - 2) ), which simplifies to
(t - 2) / t.
As t -> 2 this expression approaches zero.
*@
*********************************************
Question: How many revolutions are made by the circular helix R(t) = (sin t)i + (cos t)j + (3/4)tk in a vertical distance of 12 units.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution: if the vertical distance in 12 units and the slope of the circular helix is (3/4) for one revolution. So for one revolutions the helix rises 3 units so if rise 12 units then it would have done 4 revolutions.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):okay
------------------------------------------------
Self-critique rating:3
@&
Your idea of using the slope is good.
However the value of t changes by more than 3 as you complete a full revolution.
*@
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!