A) 12
B) 32
C) 36
D) 10
Correct Answer: D
Solution :
Given, \[x=s+3t,\,y=2s-t\] \[\Rightarrow \,\,\frac{dx}{ds}=1,\,\frac{dy}{ds}=2\] Now, \[u={{x}^{2}}+{{y}^{2}}\Rightarrow \frac{du}{ds}=2x\frac{dx}{ds}+2y\frac{dy}{ds}=2x+4y\] \[\Rightarrow \,\frac{{{d}^{2}}u}{d{{s}^{2}}}=2\left( \frac{dx}{ds} \right)+\left( \frac{dy}{ds} \right)\] \[\Rightarrow \,\frac{{{d}^{2}}u}{d{{s}^{2}}}=2\left( 1 \right)+4\left( 2 \right)=10\]You need to login to perform this action.
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