Mathematics

A curve is represented parametrically by the equation $$x=e^t \cos t$$ and $$y=e^t \sin t$$, where $$t$$ is a parameter. Then,

If $$F(t)=\int (x+y)dt$$, then the value of $$F\left (\displaystyle \frac {\pi}{2}\right )-F(0)$$ is


ANSWER

$$e^{\tfrac{\pi}{2}}$$


SOLUTION
Given,

$$F(t)=\int (x+y)dt$$

$$=\int (e^t \cos t+e^t \sin t)dt$$

$$=\dfrac{e^t\sin \left(t\right)}{2}+\dfrac{e^t\cos \left(t\right)}{2}-\dfrac{e^t\cos \left(t\right)}{2}+\dfrac{e^t\sin \left(t\right)}{2}$$

$$=e^t\sin t+c$$

Now,

$$F\left ( \dfrac{\pi}{2} \right )-F(0)=e^{\frac{\pi}{2}}\sin \dfrac{\pi}{2}-(e^0\sin 0)$$

$$=e^{\frac{\pi}{2}}(1)-1 (0)$$

$$=e^{\frac{\pi}{2}}$$
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Single Correct Medium Published on 17th 09, 2020
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