I have a question as Follows:
Let F(x,y) = ( (x^2 + y^2),(x^2 + y^2) ), and let C be the straight line segment of length 1, with an end point at the origin. Find the direction of C such that the line integral $$\int \vec F \cdot d\vec R\ $$ is a maximum, a minimum and zero. Give the direction of C and the value of the integral.
I have been stuck for hours on this ! My first would be to just use general values for (x,y), plug it into the $$ \vec r(t) = (1-t)(0,0) + t(x,y) = (xt,yt) $$ $$ d\vec r = (x,y)$$ and let $$ \vec F = (x^2 + y^2) dx + (x^2 + y^2) dy $$
and just go from there. However my mind just goes blank at this stage and I don't know if I'm on the right path...
Any help appreciated!
UPDATE:
Thanks for the advice and help everyone. I didn't find an answer by continuing along the same path as mentioned above. Still got stuck.... What I did was to find the gradient vector and that would be the greatest direction of change. then following the definition $$ \int \vec F \cdot T ds $$
T in my case would be the curve C which has length 1.By using the identity that $$ [ \vec F \cdot T ] =||F||||T||Cos(\theta ) $$ $$[ \vec F \cdot T ] =||F||(1)Cos(\theta ) $$ and length vector F is $$ || \vec F || = \sqrt{ (x^2+y^2)^2 + (x^2+y^2)^2} = \sqrt{1^2 + 1^2} = \sqrt{2} $$ Now we got $$ \sqrt{2}Cos(\theta) $$ and from here it would be easy to get the max, min and value of zero for the line integral.
Any ideas on using this approach?
Continuing the good work you began with, but changing to $\;(a,b)\;$ instead of $\;(x,y)\;$... and remember: $a^2+b^2=1\;$ (why?) :
$$\int_C\vec F\cdot d\vec r=\int_0^1(a^2t^2+b^2t^2,\,a^2t^2+b^2t^2)\cdot(a,b)dt=\int_0^1\left(a^3+ab^2+a^2b+b^3\right)t^2\,dt=$$
$$\frac13\left(a^3+ab^2+a^2b+b^3\right)=\frac13\left(a(a^2+b^2)+b(a^2+b^2)\right)=\frac13(a+b)$$
Well, now you have the easy task to find out where the above is zero, maximal and minimal, when $\;(a,b)\;$ belongs to the unit circle...