A) 1
B) 8
C) 13
D) 5
Correct Answer: C
Solution :
Let \[W=nw\]\[\Rightarrow \]\[\frac{dW}{dt}=n\frac{dw}{dt}+w.\frac{dn}{dt}\] Given :\[w={{t}^{2}}-t+2\]and\[n=2{{t}^{2}}+3\] \[\Rightarrow \]\[\frac{dw}{dt}=2r-1\]and\[\frac{dn}{dt}=4t\] \[\therefore \]Equation (1) \[\Rightarrow \]\[\frac{dw}{dt}=(2{{t}^{2}}+3)(2r-1)+({{t}^{2}}-t+2)(4t)\] Thus,\[{{\left. \frac{dw}{dt} \right|}_{t=1}}=(2+3)(2-1)+(2)(4)\] \[=5(1)+8=13\]You need to login to perform this action.
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