Answer
a) $\frac{dA}{dt}=\frac{1}{2}ab\cos\theta\frac{d\theta}{dt}$
b) $\frac{dA}{dt}=\frac{b}{2}(\sin\theta\frac{da}{dt}+a\cos\theta\frac{d\theta}{dt})$
c) $\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$
Work Step by Step
Given $A=\frac{1}{2}ab\sin\theta$
a) Take $a, b$ as constants , we have
Thus, $\dfrac{dA}{dt}=\frac{1}{2}ab\cos\theta\frac{d\theta}{dt}$
b) Now, take $b$ as a constant and differentiating $A$ w.r.t. time, we have:
$\frac{dA}{dt}=\frac{b}{2}(\sin\theta\frac{da}{dt}+a\cos\theta\frac{d\theta}{dt})=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\frac{d\sin\theta}{dt})$
or, $\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$
c) When no variables are constant and then use the triple product rule, we get:
$(uvw)'=u'vw+uv'w+uvw'$
$\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\frac{d\sin\theta}{dt})=\dfrac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$
Hence,
a) $\frac{dA}{dt}=\frac{1}{2}ab\cos\theta\frac{d\theta}{dt}$
b) $\frac{dA}{dt}=\frac{b}{2}(\sin\theta\frac{da}{dt}+a\cos\theta\frac{d\theta}{dt})$
c) $\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$
