Mathematics: WHAT IS YOU TYPE (IN PEOPLE) FREE POINTTS

WHAT IS YOU TYPE (IN PEOPLE) FREE POINTTS - id9566235, nataliaalejandradasi

Ответ:

edgytaehyung

21.01.2022

Mathematics

∫ F dr= 1

Step-by-step explanation:

The correct question is -

Given - $F = < \frac{sec^{2} x}{y} , - \frac{tan x}{y^{2} }$ and C : (0, 1) to $(\frac{\pi }{4} , 1)$

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

$d(\frac{tanx}{y }) = \frac{d}{dx}(\frac{tanx}{y}) + \frac{d}{dy}(\frac{tanx}{y})$

So,

$\frac{d}{dx}(\frac{tanx}{y}) =\frac{1}{y} \frac{d}{dx}({tanx}) \\\\ = \frac{sec^{2} x}{y} \\\frac{d}{dy}(\frac{tanx}{y}) = tanx \frac{d}{dy}(y^{-1}) \\ = -\frac{tanx}{y^{2} }$

Now,

We can see that,

$d(\frac{tanx}{y }) = \frac{sec^{2} x}{y} dx - (\frac{tanx}{y^{2} })dy$

So,

$\int\limits^{}_C {F. dr} = \int\limits^{}_C d(\frac{tanx}{y} )\\= [\frac{tanx}{y} ]\limits^{(\frac{\pi }{4} ,1)}_{(0,1)}\\= [\frac{tanx}{y} ]\limits^{}_{(\frac{\pi }{4} ,1)} - [\frac{tanx}{y} ]\limits^{}_{(0,1)}\\= \frac{tan(\frac{\pi }{4} )}{1} - \frac{tan 0}{1}\\= 1 - 0\\= 1$

∴ we get

∫ F dr= 1

Evaluate and input the result of the following integral.
∫ Fdr= ?
c
F=

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WHAT IS YOU TYPE (IN PEOPLE) FREE POINTTS

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