find the exact length of the curve.\nx = e^{t}-4t, y = 8e^{t/2}, 0 ≤ t ≤ 2

Explanation:

Step1: Find derivatives

For (x = e^{t}-4t), the derivative (\frac{dx}{dt}=e^{t}-4). For (y = 8e^{t/2}), the derivative (\frac{dy}{dt}=8\times\frac{1}{2}e^{t/2}=4e^{t/2}).

Step2: Use the arc - length formula for parametric curves

The arc - length formula for parametric curves (x = x(t)), (y = y(t)) with (a\leq t\leq b) is (L=\int_{a}^{b}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}dt). Here, (a = 0), (b = 2), ((\frac{dx}{dt})^{2}=(e^{t}-4)^{2}=e^{2t}-8e^{t}+16), ((\frac{dy}{dt})^{2}=16e^{t}). Then ((\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}=e^{2t}-8e^{t}+16 + 16e^{t}=e^{2t}+8e^{t}+16=(e^{t}+4)^{2}).

Step3: Integrate

(L=\int_{0}^{2}\sqrt{(e^{t}+4)^{2}}dt=\int_{0}^{2}(e^{t}+4)dt). Integrate term - by - term: (\int(e^{t}+4)dt=e^{t}+4t+C). Evaluate the definite integral: (\left[e^{t}+4t\right]_{0}^{2}=(e^{2}+4\times2)-(e^{0}+4\times0)). Since (e^{0}=1), we have (L=e^{2}+8 - 1=e^{2}+7).

Answer:

(e^{2}+7)