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Find the exact area of the surface obtained by

x=e^(t)-t,quad y=4e^((t)/(2)),quad0 <= t <= 1

Find the exact area of the surface obtained by $\newline$ $x=e^{t}-t, \quad y=4 e^{\frac{t}{2}}, \quad 0 \leq t \leq 1$

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Domain and range of quadratic functions: equations

Full solution

Q. Find the exact area of the surface obtained by

\newline

x=e^{t}-t, \quad y=4 e^{\frac{t}{2}}, \quad 0 \leq t \leq 1

Calculate Derivatives: Calculate the derivatives of $x$ and $y$ with respect to $t$ . $\frac{dx}{dt} = \frac{d}{dt} (e^t - t) = e^t - 1$ $\frac{dy}{dt} = \frac{d}{dt} (4e^{\frac{t}{2}}) = 2e^{\frac{t}{2}}$
Use Surface Area Formula: Use the formula for the surface area of a curve given by parametric equations $x(t)$ and $y(t)$ : $S = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$ . Here, $a = 0$ and $b = 1$ . Substitute $\frac{dx}{dt}$ and $\frac{dy}{dt}$ into the formula. $S = \int_{0}^{1} \sqrt{(e^t - 1)^2 + (2e^{\frac{t}{2}})^2} dt$
Simplify Expression: Simplify the expression under the square root. $\newline$ $S = \int_{0}^{1} \sqrt{(e^{t} - 1)^{2} + 4e^{t}} \, dt$
Recognize Perfect Square: Further simplify the expression. $\newline$ $S = \int_{0}^{1} \sqrt{e^{2t} - 2e^{t} + 1 + 4e^{t}} \, dt$ $\newline$ $S = \int_{0}^{1} \sqrt{e^{2t} + 2e^{t} + 1} \, dt$
Integrate Function: Recognize the expression under the square root as a perfect square. $\newline$ S = $\int_{0}^{1} \sqrt{(e^{t} + 1)^{2}} \, dt$ $\newline$ S = $\int_{0}^{1} (e^{t} + 1) \, dt$
Integrate Function: Recognize the expression under the square root as a perfect square. $\newline$ $S = \int_{0}^{1} \sqrt{(e^{t} + 1)^{2}} \, dt$ $\newline$ $S = \int_{0}^{1} (e^{t} + 1) \, dt$ Integrate the function from $0$ to $1$ . $\newline$ $S = [e^{t} + t]_{0}^{1}$ $\newline$ $S = (e^{1} + 1) - (e^{0} + 0)$ $\newline$ $S = e + 1 - 1$ $\newline$ $S = e$

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