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.

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

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

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

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

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