## University Calculus: Early Transcendentals (3rd Edition)

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