Answer
$\frac{dA}{dt}=\frac{1}{2}ab\cos\theta\frac{d\theta}{dt}$
$\frac{dA}{dt}=\frac{b}{2}(\sin\theta\frac{da}{dt}+a\cos\theta\frac{d\theta}{dt})$
$\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$
Work Step by Step
Given $A=\frac{1}{2}ab\sin\theta$
on taking a and b constant and differentiating the terms, we get:
$\frac{dA}{dt}=\frac{1}{2}ab\cos\theta\frac{d\theta}{dt}$
taking b as a constant and differentiating A, we get:
$\frac{dA}{dt}=\frac{b}{2}(\sin\theta\frac{da}{dt}+a\cos\theta\frac{d\theta}{dt})$
when no variables are constant and applying the triple product rule, we get:
(uvw)'=u'vw+uv'w+uvw'
$\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\frac{d\sin\theta}{dt})$
$\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$
Thus the final answer is:
$\frac{dA}{dt}=\frac{1}{2}ab\cos\theta\frac{d\theta}{dt}$
$\frac{dA}{dt}=\frac{b}{2}(\sin\theta\frac{da}{dt}+a\cos\theta\frac{d\theta}{dt})$
$\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$
