Answer:
∫ F dr= 1
c
Step-by-step explanation:
The correct question is -
Given - [tex]F = < \frac{sec^{2} x}{y} , - \frac{tan x}{y^{2} }>[/tex] and C : (0, 1) to [tex](\frac{\pi }{4} , 1)[/tex]
To find - Evaluate and input the result of the following integral.
Proof -
We know that,
dr = dx i + dy j
F = < F1, F2 >
Now, we know
[tex]d(\frac{tanx}{y }) = \frac{d}{dx}(\frac{tanx}{y}) + \frac{d}{dy}(\frac{tanx}{y})[/tex]
So,
[tex]\frac{d}{dx}(\frac{tanx}{y}) =\frac{1}{y} \frac{d}{dx}({tanx}) \\\\ = \frac{sec^{2} x}{y} \\\frac{d}{dy}(\frac{tanx}{y}) = tanx \frac{d}{dy}(y^{-1}) \\ = -\frac{tanx}{y^{2} }[/tex]
Now,
We can see that,
[tex]d(\frac{tanx}{y }) = \frac{sec^{2} x}{y} dx - (\frac{tanx}{y^{2} })dy[/tex]
So,
[tex]\int\limits^{}_C {F. dr} = \int\limits^{}_C d(\frac{tanx}{y} )\\= [\frac{tanx}{y} ]\limits^{(\frac{\pi }{4} ,1)}_{(0,1)}\\= [\frac{tanx}{y} ]\limits^{}_{(\frac{\pi }{4} ,1)} - [\frac{tanx}{y} ]\limits^{}_{(0,1)}\\= \frac{tan(\frac{\pi }{4} )}{1} - \frac{tan 0}{1}\\= 1 - 0\\= 1[/tex]
∴ we get
∫ F dr= 1
c
