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Wislar
Group Title
Please Check!
Line integral: \[\int\limits_{C}^{}x ^{2}dx+y ^{2}dy\] of the line segment from (0,2) to (4,3)
 one year ago
 one year ago
Wislar Group Title
Please Check! Line integral: \[\int\limits_{C}^{}x ^{2}dx+y ^{2}dy\] of the line segment from (0,2) to (4,3)
 one year ago
 one year ago

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Wislar Group TitleBest ResponseYou've already chosen the best response.0
Is it: \[\int\limits_{0}^{4}16t ^{2}(4)+(2+t)^{2}(1)dt\]
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
The parameterized curve is r(t)=(4t, 2+t) right?
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
@phi @amistre64 @Zarkon @TuringTest
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
id have a little reading to do to catch up on this :)
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
I think I have it down correctly, I'm just not sure if the limit is from 0 to 4
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
we need to parametrize the line from 0,2 to 4,3 right?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
4,3 02  <4,1> is the direction vector, and we can apply it to either point given so one parametric could be: x = 0 + 4t y = 2 + t
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
That's what I got! I'm not sure about the limits of integration though
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
sketch out the graph.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
the dx and dy parts in your post got me confuddled. im used to seeing a ds in it
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
here we go http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtII.aspx
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
\[\int_CPdx+Qdy=\int_CPdx+\int_CQdy\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
x = 0 + 4t dx = 4 dt y = 2 + t dy = dt
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
I have that down @amistre64 I'm just really not sure about the limits. I just used the x coordinates.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
what are the limits of t for the given line?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
when t=0 we are at (0,2) when t=? we are at (4,3)
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
I'm not sure?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
0 = 0 + 4t 2 = 2 + t t=0 4 = 0 + 4t 3 = 2 + t t = 1 so the limit of our t is 0 to 1
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
Wait how did you do that?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
we are changing x and y into functions of t correct?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
then "with respect to t" the line is from t=0 to t=1 since (0,2) is t=0; and (4,3) is t=1
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
we are simply defining all the terms with respect to t; x,y and movement
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
I have another question that asks for the line integral that consists of the arc of the circle x^2+y^2=4 from (2,0 to (0,2) I know how to find the line integral, but the limits are what I'm not sure about again
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
The parameterized curve is (2cost,2sint)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
is it asking you to define it by the line that goes from 2,0 to 0,2? or from the arc that goes from 2,0 to 0,2 ?
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
It says the arc of the circle from (2,0) to (0,2)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
x = r cos(t) y = r sin(t) and from the equation r = 2 x = 2 cos(t) dx = 2sin(t) y = 2 sin(t) dy = 2 cos(t) what is the value of t for the point x=2, y=0? what is the value of t for the point x=0, y=2?
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
cos(t)=1 and sin(t)=1?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
2 = 2 cos(t) 0 = 2 sin(t) cos(t) = 1 and sin(t) = 0 when t = 0 0 = 2 cos(t) 2 = 2 sin(t) cos(t) = 0 and sin(t) = 1 when t = pi/2
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
So it's from 0 to pi/2 right?
 one year ago

Wislar Group TitleBest ResponseYou've already chosen the best response.0
Thank you so much!!
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
youre welcome, and good luck :)
 one year ago
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