Find the length of the curve given by x= t^3, y= 2t^2, in 0≤ t≤2
"\\text{length of the curve}=\\int_a^b \\sqrt{x_t^2+y_t^2}dt\\newline\n=\\int_0^2 \\sqrt{(3t^2)^2+(4t)^2}dt\\newline\n=\\int_0^2 \\sqrt{9t^4+16t^2}dt\\newline\n=\\int_0^2 t\\sqrt{9t^2+16}dt\\newline\nPut\\space u=9t^2+16\\newline\n\\implies du=18t dt\\newline\nTherefore, \\newline\n\\int_0^2 \\sqrt{(3t^2)^2+(4t)^2}dt=\\frac{1}{18}\\int \\sqrt{u}du\\newline\n=\\frac{2}{27} {u^{3\/2}}\\newline\n=\\frac{2}{27} [(9t^2+16)^{3\/2}]_0^2\\newline\n=11.52"
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