Question

Find the line integral of f(x,y) = ye^x^2 along the curve r(t) = 4t i + 3t j -1<=t<=0

Answer 1

x(t) =4t
y(t)=3t

Answer 2

x'(t) = 4
y'(t) = 3

Answer 3

You have to compute
\[
\int_{-1}^0 f(x(t),y(t)) \sqrt{x'(t)^2+y'(t)^2}\, dt
\]

Answer 4

That is an easy integral, if you compute the integrand right

Answer 5

You seem to have left

Answer 6

Because \(t = \dfrac{x}{4} = \dfrac{y}{3} \), you are moving along \(y = \dfrac{3}{4}x\), where \(x \in [-4,0]\).

So you can also say: \(\int_C y e^{x^2} ~ ds = \int\limits_{x=-4}^0 y e^{x^2} \sqrt{1 + y'^2}~ dx\) , swapping in the x terms....

Answer 7

It is another way of doing it @IrishBoy123

Answer 8

Both methods give the same answer
\[ \bf \Large
-\frac{15}{32} \left(e^{16}-1\right)
\]

Answer 9

yes, sure !

i added it because i think is is some times forgotten that you don't always need to param, all you need for a line integral like this is an expression in a single variable

the same applies to some other aspects of vector calculus where there is the param goldrush. for example my goto book is Boas, and she seems to stick to Cartesian where poss, and maybe introduces other coordinates as a way of doing the integration, you know as a sub.

so her normal vector is taken off the gradient of the level surface, as opposed to via a paramaterisation and cross product.

different in style from, say, the v popular and v wonderful Paul's Online Maths notes where param-ing is the way to go.

i hope that makes sense, and not too jabbery :)