Find the length of the path $c(t) = (4t, 2t^{3/2})$ between $t=1$ and $t=3$.
We can use the standard arc length formula here to find this length.
Since $x'(t) = 4$ and $y'(t) = 3t^{1/2}$, we know that the integral we need to compute is $\displaystyle \int_1^3 \sqrt{4^2 + (3\sqrt{t})^2}\ dt $
This integral can be evaluated by substitution with $u = 16 + 9t$.
$\displaystyle \int_1^3 \sqrt{16 + 9t}\ dt = \frac{1}{9} \int_{25}^{43} \sqrt{u}\ du = \frac{2}{27}\left( (43)^{3/2} - (25)^{3/2} \right) = 11.627$
Find the length of the spiral $c(t) = (t \cos(t), t\sin(t))$ between $t=0$ and $t=4\pi$.
We can again apply the standard formula here, with $x'(t) = \cos(t) - t\sin(t)$ and $y'(t) = \sin(t) + t\cos(t)$.
Expanding and simplifying gives that $x'(t)^2 + y'(t)^2 = 1 + t^2$. Work this out for yourself, it relies on the fact that $\sin^2(t) + \cos^2(t) = 1$.
Thus, we need to compute $\displaystyle \int_0^{4\pi} \sqrt{1+t^2}\ dt$. This will require trigonometric substitution and then reduction formulas.
With $t = \tan{\theta}$, this integral becomes $\displaystyle \int \sec^3{\theta} d\theta$, which can be integrated by reduction formulas.
$80.82$
Find the minimum speed of the particle whose trajectory is given by $c(t) = (2t^3, t^{-2})$ with $t \geq 0.5$ for $t$ in seconds and $c(t)$ in meters.
To find the minimum speed, we need to first get an equation for the speed. How do we get this?
The speed is $\sqrt{x'(t)^2 + y'(t)^2}$, which for this curve is $\sqrt{(6t^2)^2 + (-2t^{-3})^2}$
To find the minimum of this, we can instead look for the minimum of the square of the speed, because that will be simpler. Thus, we need to find the minimum of the function $36t^4 + 4t^{-6}$.
The derivative of this function is $144t^3 - 24t^{-7}$, which is equal to zero at $t = 6^{-1/10}$
The minimum speed is the speed at $6^{-1/10}$, which is $29.3023$.
Find the surface area of the solid of revolution generated by revolving the curve $c(t) = (3t^2, 2t)$ between $t=1$ and $t=3$ around the $x$-axis.
We have the standard formula for surface area here: $\displaystyle 2\pi \int_a^b y(t) \sqrt{x'(t)^2 + y'(t)^2}\ dt$.
Plugging in the different components gives that we need to compute $\displaystyle 2\pi \int_1^3 2t \sqrt{36t^2 + 4}\ dt$.
This can be solved by substitution, using $u = 4 + 36t^2$, so that $du = 72t\ dt$.
The integral then becomes $\displaystyle 2 \pi \int_{40}^{328} \frac{1}{36} \sqrt{u}\ du$.
$\displaystyle \frac{\pi}{27} \left( (328)^{3/2} - (40)^{3/2} \right) = 661.75$
Find the length of the path $c(t) = (t^3 + 1, t^2 - 3)$ between $t=0$ and $t=4$.
Find the surface area of the solid of revolution generated by revolving the curve $c(t) = (\sin^2(t), \cos^2(t)$ between $t=0$ and $t = \pi/2$ around the $x$-axis.