Write the parametric equation $x = \frac{1}{1+t}$ and $y = te^t$ in the form $y = f(x)$ by eliminating the parameter. Where is the curve at $t=0$? What about at $t=6$?
To do this, we need to solve out for $t$ in one equation and plug this into the other one. One of these equations is much easier to solve out for $t$ than the other one.
Solving out for $t$ in the $x$ equation gives that $t = \frac{1}{x} - 1$. This can then be plugged into the $y$ equation.
This gives $\displaystyle \left(\frac{1}{x} - 1\right) e^{\left(\frac{1}{x} - 1\right)}$ for the function.
For the values at certain times, we can plug those into the parametric equations.
$\displaystyle y = \left(\frac{1}{x} - 1\right) e^{\left(\frac{1}{x} - 1\right)}$. At $t=0$, the curve is at the point $(1,0)$. At $t=6$, the curve is at $(1/7, 6e^6)$.
A particle follows the trajectory $x(t) = 6t - 5$, $y(t) = 10 + 3t - t^2$, with $t$ in seconds and $x$ and $y$ in meters. What is the maximum height of the particle? When does it hit the ground and how far from the origin does it land?
What does it mean for the particle to be at its maximum height? It means that the function $y(t)$ is at a maximum. How do we find this?
Take the derivative of the $y(t)$ equation, set to zero, and then plug this back in to find the maximum.
What does it mean for the particle to hit the ground? It means that $y(t) = 0$.
This gives the value of $t = 5$, since that is the positive solution to $y(t) = 0$, and then that can be plugged into the $x$ equation.
The maximum height is at $t = 3/2$, and is $49/4$. The particle hits the ground at $t=5$, at position $25$.
Find a parametrization for the curve $3y = 7x^2 - 2x$.
We need to introduce a parameter here to make this happen. Since it is almost already written as $y = f(x)$, we have an easy choice for parameter.
Set $x = t$, and then write $y$ in terms of $t$ using that.
The parametrization we get here is $x=t$, $\displaystyle y = \frac{7t^2 - 2t}{3}$.
Find a parametrization $c(t)$ for the curve $x^2 + y^2 = 4$, satisfying $c(0) = (-1, \sqrt{3})$.
The curve here is a circle. How can we write a parametrization for a circle?
To do this, we want to use sine and cosine, so one possible parametrization could be $c(t) = (2\cos(t), 2 \sin(t))$.
However, for this parametrization $c(0) = (2,0)$, not the value we need. How can we modify our parametrization to fit this desired condition? The easiest way is by shifting the argument $t$.
We know that $\displaystyle c\left(\frac{2\pi}{3}\right) = \left(-1, \sqrt{3}\right)$ because of the values of sine and cosine. How can we use this fact to give us a new parametrization that works and has this value at $0$?
$\displaystyle c(t) = \left( 2\cos\left( t + \frac{2\pi}{3}\right), 2\sin\left(t + \frac{2\pi}{3}\right) \right)$
Find the equation of the tangent line to the parametric curve $x = \sec{\theta}$, $y = \tan{\theta}$ at the point $(2, \sqrt{3})$.
The first thing we need to figure out is what value of $\theta$ corresponds to the point $(2, \sqrt{3})$, so we need to figure out what angle $\theta$ has $\sec(\theta) = 2$ and $\tan{\theta} = \sqrt{3}$.
We find that $\theta = \frac{\pi}{3}$. Then, we can use our standard formula for derivatives to figure out the slope of the tangent line.
Since $\displaystyle \frac{dy}{dx} = \frac{y'(\theta)}{x'(\theta)}$, we compute that $\frac{dy}{dx} = \frac{\sec{\theta}\tan{\theta}}{\sec^2{\theta}} = \frac{\tan{\theta}}{\sec{\theta}} = \sin{\theta}$.
Thus, the slope of the tangent line should be $\sin(\pi/3) = \frac{\sqrt{3}}{2}$. This will let us write the equation of the line.
$\displaystyle y - \sqrt{3} = \frac{\sqrt{3}}{2}\left( x - 2\right)$
Find the area under the graph of $c(t) = \left(\ln{t}, 2 - t\right)$ between $t=1$ and $t=2$
We can apply the standard formula $\displaystyle \int_a^b y(t) x'(t)\ dt$ to find this area.
We use $y(t) = 2-t$ and $x'(t) = \frac{1}{t}$ to solve this problem.
We need to integrate $\displaystyle \int_1^2 (2-t)\frac{1}{t}\ dt = \int_1^2 \frac{2}{t} - 1 \ dt$.
The antiderivative here is $2 \ln{t} - t$ and then we can plug in $1$ and $2$.
$2 \ln{2} - 2 - (2 \ln{1} - 1) = 2\ln{2} - 1$
Find a parametrization for the ellipse $\displaystyle \left(\frac{x}{3}\right)^2 + \left(\frac{y}{4}\right)^2 = 1$.
Find all points where the tangent lines to the curve $c(t) = (\frac{t^3}{3} + t^2 - 2t - 1 , 2t^3 - 4t^2 + 4t + 3)$ have slope 2.