The derivative of a parametric function is given by the following:

$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

So, we must find the derivative of each function at the t given:

$\displaystyle y'=e^t+te^t$, $\displaystyle x'=2t\cos(t)-t^2\sin(t)+3$

The derivatives were found using the following rules:

$\displaystyle \frac{d}{dx}(e^x)=e^x$, $\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}$, 

$\displaystyle \frac{d}{dx}\sin(x)=\cos(x)$, 

$\displaystyle \frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

Next, plug in the given $\displaystyle t=0$ into each derivative function:

$\displaystyle y'(0)=1$, $\displaystyle x'(0)=3$

Finally, divide $\displaystyle \frac{y'}{x'}$ to get a final answer of $\displaystyle \frac{1}{3}$.

Finding $\displaystyle \frac{dy}{dx}$ of a parametric equation can be given by this formula: 

$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

So we must find $\displaystyle \frac{dy}{dt}$ and $\displaystyle \frac{dx}{dt}$ for when $\displaystyle t = 2$. 

$\displaystyle \frac{dy}{dt} = 10t + 3$ and $\displaystyle \frac{dx}{dt} = \frac{1}{t}$ and so 

$\displaystyle \frac{dy}{dx} = 10t^{2} + 3t$.

When you plug in  $\displaystyle t = 2$ you get your answer $\displaystyle 46$.

This parametric equation is described as the sum of three vectors.  To find the derivative of a parametric equation, you must find the derivative of each vector, or if

$\displaystyle \mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+x(t)\mathbf{k}$  then  $\displaystyle \mathbf{r'}(t)=x'(t)\mathbf{i}+y'(t)\mathbf{j}+x'(t)\mathbf{k}$

The derivative of the first vector is found using the power rule, 

$\displaystyle \frac{d}{dt}(t^{c})=ct^{c-1}$ where $\displaystyle c$ is a constant.

The derivative of the second vector is found using the natural logarithm rule,

$\displaystyle \frac{d}{dt}[ln(t)]=\frac{1}{t}$.

The derivative of the third vector is found using one of the trigonmetric rules,

$\displaystyle \frac{d}{dt}[sin(t)]=cos(t)$.

In this case:

$\displaystyle \mathbf{r'}(t)=\frac{d}{dt}(t^{3})\mathbf{i}+3\frac{d}{dt}[ln(t)]\mathbf{j}+5\frac{d}{dt}[sin(t)]\mathbf{k}$

$\displaystyle =3t^{2}\mathbf{i}+\frac{3}{t}\mathbf{j}+5cos(t)\mathbf{k}$

This parametric equation is described as the sum of three vectors.  To find the derivative of a parametric equation, you must find the derivative of each vector, or if

$\displaystyle \mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+x(t)\mathbf{k}$,  then  $\displaystyle \mathbf{r'}(t)=x'(t)\mathbf{i}+y'(t)\mathbf{j}+x'(t)\mathbf{k}$

The derivative of the first vector is found using the power rule, 

$\displaystyle \frac{d}{dt}(t)=1$.

The derivative of the second vector is found using the exponential rule,

$\displaystyle \frac{d}{dt}(e^{t})=e^{t}$.

The derivative of the third vector is found using one of the trigonmetric rules,

$\displaystyle \frac{d}{dt}[sin(at)]=acos(at)$, where $\displaystyle a$ is a constant.

In this case:

 $\displaystyle \mathbf{r'}(t)=\frac{d}{dt}(t)\mathbf{i}+3\frac{d}{dt}(e^{t})\mathbf{j}+\frac{d}{dt}[sin(\pi t)]\mathbf{k}$

$\displaystyle =\mathbf{i}+3e^{t}\mathbf{j}+\pi cos(\pi t)\mathbf{k}$

This parametric equation is described as the sum of three vectors.  To find the derivative of a parametric equation, you must find the derivative of each vector, or if

$\displaystyle \mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+x(t)\mathbf{k}$,  then  $\displaystyle \mathbf{r'}(t)=x'(t)\mathbf{i}+y'(t)\mathbf{j}+x'(t)\mathbf{k}$

The derivative of the first and second vectors are found using the following trigonometric rules, 

$\displaystyle \frac{d}{dt}[cos(at)]=-asin(at)$ and  $\displaystyle \frac{d}{dt}[sin(bt)]=bcos(bt)$,

where $\displaystyle a$ and $\displaystyle b$ are constants.

In this case:

$\displaystyle \mathbf{r'}(t)=2\frac{d}{dt}[cos(5t)]\mathbf{i}+5\frac{d}{dt}[sin(7t)]\mathbf{j}$

$\displaystyle =[2*(-sin(5t))*5]\mathbf{i}+[5*cos(7t)*7]\mathbf{j}$

$\displaystyle =-10sin(5t)\mathbf{i}+35cos(7t)\mathbf{j}$

If $\displaystyle x=x(t)$ and $\displaystyle y=y(t)$, then we can use the chain rule to define $\displaystyle \frac{dy}{dx}$ as 

$\displaystyle \frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}$.  

We then use the following trigonometric rules, 

$\displaystyle \frac{d}{dt}[cos(at)]=-asin(at)$ and  $\displaystyle \frac{d}{dt}[sin(bt)]=bcos(bt)$,

where $\displaystyle a$ and $\displaystyle b$ are constants.

In this case:

 $\displaystyle \frac{dy}{dt}=\frac{d}{dt}(cos(7t))=-7sin(7t)$,

 and

 $\displaystyle \frac{dx}{dt}=\frac{d}{dt}(sin(7t))=7cos(7t)$,

therefore 

$\displaystyle \frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{-7sin(7t)}{7cos(7t)}=-tan(7t)$.

The formula for arc length of a parametric curve in space is $\displaystyle \int_{a}^{b} \sqrt{(\frac{d\mathbf{v}_1}{dt})^2 +(\frac{d\mathbf{v}_2}{dt})^2 +(\frac{d\mathbf{v}_3}{dt})^2} dt$ for $\displaystyle a < t< b$.

Taking derivatives of each of the vector function components and substituting the values into this formula gives

$\displaystyle \int_{0}^{1} \sqrt{1 + 2e^{2t} + e^{4t}} dt$

We need to recognize that underneath the square root we have a perfect square, and we can write it as

$\displaystyle \int_{0}^{1} \sqrt{(e^{2t} + 1)^2} dt$

Solving this we get

$\displaystyle \int_{0}^{1} e^{2t}+1 dt = [\frac{1}{2}e^{2t} +t]_0^1$ $\displaystyle = \bigg(\frac{1}{2}e^2+1\bigg)-\bigg(\frac{1}{2}\bigg) = \frac{1}{2}(e^2+1)$

The correct answer is $\displaystyle 1$.

We use the equation

$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{1}{t+1}+2t+0}{e^t}$

But we need a value for $\displaystyle t$ to substitute into our derivative. We can obtain such a $\displaystyle t$ by setting $\displaystyle x(t) =1, y(t) =2$ as our given point suggests.

$\displaystyle 1 = e^t$ $\displaystyle \Rightarrow t =0$
$\displaystyle 2 = ln(t+1)+t^2+2 \Rightarrow t=0$
Since our values of $\displaystyle t$ match, $\displaystyle t=0$ is our correcct value. Substituting this into the derivative and simplifying gives us our answer of $\displaystyle 1.$

When the problem asks us to eliminate the parametric, that means we want to somehow get rid of our variable t and be left with an equation that is only in terms of x and y. While the equation for x is a polynomial, making it more difficult to solve for t, we can see that the equation for y can easily be solved for t:

$\displaystyle y=\frac{3}{2}t-6$

$\displaystyle t=\frac{2}{3}y+4$

Now that we have an expression for t that is only in terms of y, we can plug this into our equation for x and simplify, and we will be left with an equation that is only in terms of x and y:

$\displaystyle x=t^2+3t+5=(\frac{2}{3}y+4)^2+3(\frac{2}{3}y+4)+5$

$\displaystyle x=\frac{4}{9}y^2+\frac{16}{3}y+16+2y+12+5$

$\displaystyle x=\frac{4}{9}y^2+\frac{22}{3}y+33$

Write the arc length formula for parametric curves.

$\displaystyle L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\:dt$

Find the derivatives.  The bounds are given in the problem statement.

$\displaystyle \frac{dx}{dt}=-sin(t)$

$\displaystyle \frac{dy}{dt}=cos(t)$

$\displaystyle L=\int_{0}^{\pi}\sqrt{(-sin(t))^2+(cos(t))^2}\:dt$

$\displaystyle L=\int_{0}^{\pi}\sqrt{sin(t)^2+cos(t)^2}\:dt$

$\displaystyle L=\int_{0}^{\pi}1\:dt =[t]_{0}^{\pi}= \pi$