Answers: 2
![\displaystyle \frac{d}{dt}[c^2=a^2+b^2-2ab\cos(\theta)]\\\\\\\Rightarrow2c\frac{dc}{dt}=\frac{da}{dt}(2a-2b\cos(\theta))+\frac{db}{dt}(2b-2a\cos(\theta))+2ab(\sin(\theta)\frac{d\theta}{dt})](/tpl/images/0921/9937/31613.png)
![\displaystyle\frac{d}{dt}[c^2]=\frac{d}{dt}[a^2+b^2-2ab\cos(\theta)]](/tpl/images/0921/9937/36d53.png)
![\displaystyle \frac{d}{dt}[c^2]=\frac{d}{dt}[a^2]+\frac{d}{dt}[b^2]-\frac{d}{dt}[2ab\cos(\theta)]](/tpl/images/0921/9937/b9aad.png)
![\displaystyle 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}-2\frac{d}{dt}[ab\cos(\theta)]](/tpl/images/0921/9937/b74a0.png)
![\displaystyle \frac{d}{dt}[ab\cos(\theta)]](/tpl/images/0921/9937/6378a.png)
![\displaystyle u'=\frac{d}{dt}[ab]\text{ and } v'=\frac{d}{dt}[\cos(\theta)]](/tpl/images/0921/9937/458c4.png)
![\displaystyle u^\prime=\frac{d}{dt}[ab]](/tpl/images/0921/9937/ae444.png)
![\displaystyle u_2^\prime=\frac{d}{dt}[a]\text{ and }v_2^\prime=\frac{d}{dt}[b]](/tpl/images/0921/9937/b0676.png)
![\displaystyle u{^\prime}=\frac{d}{dt}[ab]=u_2^\prime v_2+u_2v_2^\prime=\frac{da}{dt}b+a\frac{db}{dt}](/tpl/images/0921/9937/de28a.png)
![\displaystyle \frac{d}{dt}[ab\cos(\theta)]=u^\prime v+uv^\prime\\\\ =(\frac{da}{dt}b+a\frac{db}{dt})\cos(\theta)-(ab)(\sin(\theta)\frac{d\theta}{dt})](/tpl/images/0921/9937/0dc8f.png)
![\displaystyle 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}-2\Big[(\frac{da}{dt}b+a\frac{db}{dt})\cos(\theta)-(ab)(\sin(\theta)\frac{d\theta}{dt})\Big]](/tpl/images/0921/9937/8691c.png)
![\displaystyle 2c\frac{dc}{dt}=\big[2a\frac{da}{dt}-2\frac{da}{dt}b\cos(\theta)\big]+\big[2b\frac{db}{dt}-2a\frac{db}{dt}\cos(\theta)\big]+2ab(\sin(\theta)\frac{d\theta}{dt})](/tpl/images/0921/9937/d6b6d.png)
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