To find the length of a curve given by parametric equations \( x = e^t - t \) and \( y = 4e^{t/2} \) over the interval \( -8 \leq t \leq 3 \), we can use the formula for the arc length of a curve in parametric form. This formula is given by:
\[ L = \int_{a}^{b} \!\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \,dt \]
where \( x(t) \) and \( y(t) \) are the parametric equations for the curve and \( a \) and \( b \) are the bounds of the interval over which we want to determine the length.
First, compute the derivatives of \( x(t) \) and \( y(t) \) with respect to \( t \):
\( x(t) = e^t - t \)
\( \frac{dx}{dt} = \frac{d}{dt}(e^t - t) = e^t - 1 \)
\( y(t) = 4e^{t/2} \)
\( \frac{dy}{dt} = \frac{d}{dt}(4e^{t/2}) = 2e^{t/2} \)
Now, plug the derivatives back into the arc length formula:
\[ L = \int_{-8}^{3} \!\sqrt{(e^t - 1)^2 + (2e^{t/2})^2} \,dt \]
Let's simplify the integrand inside the square root:
\[ (e^t - 1)^2 + (2e^{t/2})^2 = e^{2t} - 2e^t + 1 + 4e^t \]
Combine like terms:
\[ e^{2t} - 2e^t + 1 + 4e^t = e^{2t} + 2e^t + 1 \]
Now, the integral becomes:
\[ L = \int_{-8}^{3} \!\sqrt{e^{2t} + 2e^t + 1} \,dt \]
We see that this is of the form \( e^t + 1 \) which simplifies as follows:
\[ L = \int_{-8}^{3} (e^t + 1) \,dt \]
This integral can now be computed directly:
\[ L = [e^t + t]_{-8}^{3} \]
\[ L = (e^3 + 3) - (e^{-8} - 8) \]
Now, we simply need to compute the values and subtract:
\[ e^3 \approx 20.0855 \]
\[ e^{-8} \approx 0.0003 \]
\[ L \approx (20.0855 + 3) - (0.0003 - 8) \]
\[ L \approx 23.0855 + 7.9997 \]
\[ L \approx 31.0852 \]
So, the length of the curve from \( t = -8 \) to \( t = 3 \) is approximately \( 31.0852 \) units.
