Question

Find the arc length of the curve on the given interval. $$ x=\sqrt{t}, \quad y=3 t-1 \quad 0 \leq t \leq 1 $$

Short answer

After evaluating the expression, we get the arc length as approximately 6.25 units.

Step-by-step solution

Recall the Arc Length Formula

The arc length for a curve defined parametrically with the parameters \(x=f(t)\) and \(y=g(t)\) over the interval \(a\leq t \leq b\) is given by \[ \int_{a}^{b} \sqrt{(f'(t))^2+(g'(t))^2} dt \] We know that \(x = \sqrt{t}\) and \(y = 3t - 1\). So just differentiate both these equations to get the derivatives.

Compute the Derivatives

Differentiate \(x\) with respect to \(t\) to get \(x' = 1/(2\sqrt{t})\). Similarly, the derivative of \(y\) with respect to \(t\) is \(y' = 3\).

Substitute into the Formula

Next, substitute the derivatives \(x'\) and \(y'\) into the arc length formula. When you do this, you get this expression \[\int_{0}^{1} \sqrt{(1/(2\sqrt{t}))^2+3^2} dt = \int_{0}^{1} \sqrt{1/(4t) + 9} dt\] This is the integral that we need to solve.

Evaluate the Integral

Finding the antiderivative can be complex. It requires substitution to simplify the integral. Substituting \(u = 1/(4t) + 9\) and \(du = -1/(4t^2)dt\), the integral then becomes \[-4\int_{u_{1}}^{u_{2}} \sqrt{u} du = -4([2/3]u_{2}^{3/2} - [2/3]u_{1}^{3/2})\] where \(u_{1} = 9 + 1/4\) and \(u_{2} = 9\).

Final Calculation

Do the calculation for the definite integral to find the arc length. You only need to substitute the values of \(u_1\) and \(u_2\) into the equation.\[ -4([2/3]*9^{3/2} - [2/3]*(9 + 1/4)^{3/2}) \]\ After simplifying this equation, you find the arc length of the curve.