A) 12
B) 32
C) 36
D) 10
Correct Answer: D
Solution :
\[u={{x}^{2}}+{{y}^{2}},\] \[x=s+3t,\] \[y=2s-t\] Now \[\frac{dx}{ds}=1,\] \[\frac{dy}{ds}=2\] .....(i) \[\frac{{{d}^{2}}x}{d{{s}^{2}}}=0,\] \[\frac{{{d}^{2}}y}{d{{s}^{2}}}=0\] ......(ii) Now \[u={{x}^{2}}+{{y}^{2}}\], \[\frac{du}{ds}=2x\,.\,\frac{dx}{ds}+2y\,.\,\frac{dy}{ds}\] \[\frac{{{d}^{2}}u}{d{{s}^{2}}}=2{{\left( \frac{dx}{ds} \right)}^{2}}+2x\frac{{{d}^{2}}x}{d{{s}^{2}}}+2{{\left( \frac{dy}{ds} \right)}^{2}}+2y\left( \frac{{{d}^{2}}y}{d{{s}^{2}}} \right)\] From (i) and (ii), \[\frac{{{d}^{2}}u}{d{{s}^{2}}}=2\times 1+0+2\times 4+0=10\].
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