Question

Let $x=f^{\prime \prime }\left(t\right)\cos t+f\left(t\right)\sin t$ and $y=-f^{\prime \prime }\left(t\right)\sin t+f^{\prime }\left(t\right)\cos t.$ Then $\int {[{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2]}^{\frac{1}{2}}dt$ equals

• A. $f^{\prime }\left(t\right)+f^{\prime \prime }\left(t\right)+c$
• B. $f^{\prime \prime }\left(t\right)+f^{\prime \prime \prime }\left(t\right)+c$
• C. $f\left(t\right)+f^{\prime \prime }\left(t\right)+c$
• D. $f\left(t\right)-f^{\prime \prime }\left(t\right)+c$

Answer 1

The correct option is C $f\left(t\right)+f^{\prime \prime }\left(t\right)+c$

$x=f^{\prime \prime }\left(t\right)\cos t+f\left(t\right)\sin t$ and $y=-f^{\prime \prime }\left(t\right)\sin t+f^{\prime }\left(t\right)\cos t.$

$\frac{dx}{dt}=-f^{\prime \prime }\left(t\right)\sin t+f^{\prime \prime \prime }\left(t\right)\cos t+f^{\prime }\left(t\right)\cos t+f^{\prime \prime }\left(t\right)\sin t=\cos t\left(f^{\prime }\right(t)+f^{\prime \prime \prime }(t\left)\right)$

$\frac{dy}{dt}=-f^{\prime \prime }\left(t\right)\cos t-f^{\prime \prime \prime }\left(t\right)\sin t-f^{\prime }\left(t\right)\sin t+f^{\prime \prime }\left(t\right)\cos t=-\sin t\left(f^{\prime }\right(t)+f^{\prime \prime \prime }(t\left)\right)$

$(\frac{dx}{dt}{)}^2={\cos }^{2}t(f^{\prime }\left(t\right)+f^{\prime \prime \prime }\left(t\right){)}^2$

$(\frac{dy}{dt}{)}^2={\sin }^{2}t(f^{\prime }\left(t\right)+f^{\prime \prime \prime }\left(t\right){)}^2$

$(\frac{dx}{dt}{)}^2+(\frac{dy}{dt}{)}^2=\left(f^{\prime }\right(t)+f^{\prime \prime \prime }(t){)}^2$

$\Rightarrow \int \left[\right(\frac{dx}{dt}{)}^2+(\frac{dy}{dt}{)}^2{]}^{\frac{1}{2}}dt$

$=\int \left(f^{\prime }\right(t)+f^{\prime \prime \prime }(t\left)\right)dt$

$=f\left(t\right)+f^{\prime \prime }\left(t\right)+c$