Multivariable Calculus, Line Integral

Click For Summary
To calculate the line integral of the vector field F given by F = (x, y) / √(1 - x² - y²) along the path C defined by (cos(t), sin(t)) / (1 + e^t), one should first express F in terms of the path's parameterization. This involves breaking down F into its x and y components, F_x and F_y, and substituting the parameterized coordinates x(t) and y(t). Next, compute the derivatives x'(t) and y'(t) and substitute these into the integral formula. The integral then simplifies to ∫ from 0 to ∞ of (F_x(x(t), y(t))x'(t) + F_y(x(t), y(t))y'(t)) dt, where careful algebra and trigonometric identities may lead to further simplifications. Completing these steps will yield the desired result for the line integral.
tompenny
Messages
15
Reaction score
3
Homework Statement
Calculate a line integral
Relevant Equations
$$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$
The vector field F which is given by $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$

And the line integral $$ \int_{C} F \cdot dr $$C is the path of $$\dfrac{\ (\cos (t), \sin (t))}{ 1+ e^t}$$ , and $$0 ≤ t < \infty $$

How do I calculate this? Anyone got a tip/hint? many thanks
 
  • Like
Likes Delta2
Physics news on Phys.org
tompenny said:
Homework Statement:: Calculate a line integral
Relevant Equations:: $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$

The vector field F which is given by $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$

And the line integral $$ \int_{C} F \cdot dr $$C is the path of $$\dfrac{\ (\cos (t), \sin (t))}{ 1+ e^t}$$ , and $$0 ≤ t < \infty $$

How do I calculate this? Anyone got a tip/hint? many thanks
Remember, ##\int_C \vec F \cdot d\vec r = \int_t \vec F(\vec r(t))\cdot \vec r'(t)~dt##. So the obvious hint is plug the formulas in and see what happens. Then come back to show us where you are stuck. There is no substitute for getting your hands dirty.
 
  • Like
Likes Delta2 and etotheipi
My advice would be to split ##\mathbf{F}## as well the path ##C## into x and y components. For example it will be $$F_x(x,y)=\frac{x}{\sqrt{1-x^2-y^2}}$$ $$F_y(x,y)=\frac{y}{\sqrt{1-x^2-y^2}}$$while for the path C it will be $$x(t)=\frac{\cos t}{1+e^t}$$ $$y(t)=\frac{\sin t}{1+e^t}$$. Then calculate $$x'(t)$$ and $$y'(t)$$ (derivatives of x(t),y(t)with respect to t). Then combine the whole thing to $$\int_0^{\infty} (F_x(x(t),y(t))x'(t)+F_y(x(t),y(t))y'(t))dt$$.
This might look a bit scary but i think if you do all the calculations correctly and the algebra correctly some simplifications will occur (by the use of trigonometric identities).
 
  • Love
Likes etotheipi
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
1K
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K