## Thomas' Calculus 13th Edition

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) $\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$
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) $\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$