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Line integral, vector field

Problem is attached.

To do the line integral, I need to find F(r(t)), but I don't understand how to express it. For example I looked at the online notes provided here: http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsVectorFields.aspx

I don't understand how F(r(t)) was derived in the first example. I know how to do the rest.

vector.jpg
(edited 7 years ago)
Original post by blah3210
Problem is attached.

To do the line integral, I need to find F(r(t)), but I don't understand how to express it. For example I looked at the online notes provided here: http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsVectorFields.aspx

I don't understand how F(r(t)) was derived in the first example. I know how to do the rest.

vector.jpg

Express it as a vector(either in column, row or i, j, k). F is a function taking a 3d vector to a 3d vector. In this case, F((xyz))=(2sinxcosx02z)\vec{F}(\begin{pmatrix}x\\ y\\ z \end{pmatrix}) = \begin{pmatrix}2\sin x \cos x\\0\\2z\end{pmatrix}.

A general point on the line is r(t)=(ttt2)\vec{r}(t) = \begin{pmatrix}t \\ t \\ t^2 \end{pmatrix}, so F(r(t))=(2sintcost02t2)\vec{F}(\vec{r}(t)) = \begin{pmatrix}2\sin t\cos t\\ 0 \\ 2t^2 \end{pmatrix}.
Original post by blah3210
Problem is attached.

To do the line integral, I need to find F(r(t)), but I don't understand how to express it. For example I looked at the online notes provided here: http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsVectorFields.aspx

I don't understand how F(r(t)) was derived in the first example. I know how to do the rest.

vector.jpg


To find F(r(t)) you need to sub in for each of the components for example if F=(2x,x-y,0) and the path parametrised by r(t)=(x(t),y(t),z(t))=(t,2t,t^2) (for some interval of t) then whenever you see an x you need to sub in for the x component of the path so the 2x of the vector field becomes 2t since x(t)=t. Similarly for the other components and we get F(r(t))=(2t,t-2t,0)=(2t,-t,0).

Hopefully you can see how it works for your example now.
(edited 7 years ago)

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