Question

The derivative of $e^{ax} \cos\,bx$ with respect to $x$ is $re^{ax}$ $\cos\, (bx-Tan^{-1} \frac {b}{a})$ When $a >0, b > 0$,the value of $r$ is ............

• A. $ab$
• B. $a+b$
• C. $\sqrt {a^2+b^2}$
• D. $\frac {1} {\sqrt {ab}}$

Answer 1

The Correct Option is C

Given, $\frac{d}{d x}\left(e^{a x} \cos b x\right)=r e^{a x} \cos (b x+\alpha)$, then
$r=?$
Let $y=e^{a x} \cdot \cos b x$
Let $y=e^{a x} \cdot \cos b x$
$\frac{d y}{d x}=a e^{a x} \cdot \cos b x-b e^{a x} \cdot \sin b x$
$\frac{d y}{d x}=e^{a x}(a \cos b x-b \sin b x)$
$a=r \cos \alpha
b=r \sin \alpha\}$...(i)
Then, $\frac{d y}{d x} =e^{a x} \cdot r\{\cos b x \cdot \cos \alpha-\sin b x \cdot \sin \alpha\}$
$\frac{d y}{d x}=e^{a x} \cdot r \cos (b x+\alpha)$...(ii)
Where, $\tan \alpha=\frac{b}{a} \Rightarrow \alpha=\tan ^{-1}\left(\frac{b}{a}\right)$
and $r^{2}\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)=a^{2}+b^{2}$
$r^{2}=a^{2}+b^{2}$
$r=\sqrt{a^{2}+b^{2}}$