Answer: The arc length from t = 2 to t = 4 is \(240 + 4\ln 2\).
For a parametric curve, the arc length from \(t=2\) to \(t=4\) is
\(s=\int_2^4 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).
Differentiate the given equations:
\(x=t^4-4\ln t \implies \frac{dx}{dt}=4t^3-\frac{4}{t}\),
\(y=4t^2 \implies \frac{dy}{dt}=8t\).
So
\(s=\int_2^4 \sqrt{\left(4t^3-\frac{4}{t}\right)^2+(8t)^2}\,dt\).
Simplify the expression inside the square root:
\(\left(4t^3-\frac{4}{t}\right)^2+64t^2 = 16t^6-32t^2+\frac{16}{t^2}+64t^2 = 16t^6+32t^2+\frac{16}{t^2}\).
This is a perfect square:
\(16t^6+32t^2+\frac{16}{t^2} = \left(4t^3+\frac{4}{t}\right)^2\).
Since \(t\gt 0\), the square root is
\(\sqrt{\left(4t^3-\frac{4}{t}\right)^2+(8t)^2}=4t^3+\frac{4}{t}\).
Hence
\(s=\int_2^4 \left(4t^3+\frac{4}{t}\right)\,dt = \left[t^4+4\ln t\right]_2^4\).
Now evaluate:
\(s=(4^4+4\ln 4)-(2^4+4\ln 2)=256-16+4\ln 4-4\ln 2\).
Since \(\ln 4=2\ln 2\),
\(s=240+8\ln 2-4\ln 2=240+4\ln 2\).
