In order to find the arc length, we must use the arc length formula for parametric curves:

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

Given  \(\displaystyle x=7t+9\) and \(\displaystyle y=t-1\), we can use using the Power Rule

 for all , to derive 

\(\displaystyle \frac{dx}{dt}=(1)7t^{1-1}+(0)9=7\) and \(\displaystyle \frac{dy}{dt}=(1)t^{1-1}-(0)1=1\).

Plugging these values and our boundary values for \(\displaystyle t\) into the arc length equation, we get:

\(\displaystyle L=\int_{0}^{1}\sqrt{(7)^{2}+(1)^{2}}dt\)

\(\displaystyle L=\int_{0}^{1}(\sqrt{49+1})dt\)

\(\displaystyle L=\int_{0}^{1}(\sqrt{50})dt\)

\(\displaystyle L=\int_{0}^{1}(\sqrt{25\times2})dt\)

\(\displaystyle L=\int_{0}^{1}(5\sqrt{2})dt\)

Now, using the Power Rule for Integrals \(\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}\) for all \(\displaystyle n\neq0\), we can determine that:

\(\displaystyle L=(5t\sqrt{2})_{0}^{1}\textrm{}\)

\(\displaystyle L=5(1)\sqrt{2}-5(0)\sqrt{2}\)

\(\displaystyle L=5\sqrt{2}\)

In order to find the arc length, we must use the arc length formula for parametric curves:

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

Given  \(\displaystyle x=5t+2\) and \(\displaystyle y=7t+10\),we can use using the Power Rule
for all , to derive

\(\displaystyle \frac{dx}{dt}=(1)5t^{1-1}+(0)2=5\) and 

\(\displaystyle \frac{dy}{dt}=(1)7t^{1-1}+(0)10=7\) .

Plugging these values and our boundary values for into the arc length equation, we get:

\(\displaystyle L=\int_{0}^{2}\sqrt{(5)^{2}+(7)^{2}}dt\)

\(\displaystyle L=\int_{0}^{2}(\sqrt{25+49})dt\)

\(\displaystyle L=\int_{0}^{2}(\sqrt{74})dt\)

Now, using the Power Rule for Integrals

for all ,

we can determine that:

\(\displaystyle L=[t\sqrt{74}]_{0}^{2}\textrm{}dt\)

\(\displaystyle L=2\sqrt{74}-0\sqrt{74}\)

\(\displaystyle L=2\sqrt{74}\)

In order to find the arc length, we must use the arc length formula for parametric curves:

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

Given  \(\displaystyle x=6t-4\) and \(\displaystyle y=4t+6\), we can use using the Power Rule

 for all , to derive 

\(\displaystyle \frac{dx}{dt}=(1)6t^{1-1}-(0)4=6\) and 

\(\displaystyle \frac{dy}{dt}=(1)4t^{1-1}+(0)6=4\).

Plugging these values and our boundary values for  into the arc length equation, we get:

\(\displaystyle L=\int_{0}^{3}\sqrt{(6)^{2}+(4)^{2}}dt\)

\(\displaystyle L=\int_{0}^{3}(\sqrt{36+16})dt\)

\(\displaystyle L=\int_{0}^{3}(\sqrt{52})dt\)

\(\displaystyle L=\int_{0}^{3}(\sqrt{13\times4})dt\)

\(\displaystyle L=\int_{0}^{3}(2\sqrt{13})dt\)

Now, using the Power Rule for Integrals

 for all ,

we can determine that:

\(\displaystyle L=[2t\sqrt{13}]_{0}^{3}\textrm{}dt\)

\(\displaystyle L=2(3)\sqrt{13}-2(0)\sqrt{13}\)

\(\displaystyle L=6\sqrt{13}\)

In order to find the arc length, we must use the arc length formula for parametric curves:

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

Given  \(\displaystyle x=t+10\) and \(\displaystyle y=2t-9\), we can use using the Power Rule

 for all , to derive 

\(\displaystyle \frac{dx}{dt}=(1)t^{1-1}+(0)10=1\) and 

\(\displaystyle \frac{dy}{dt}=(1)2t^{1-1}-(0)9=2\).

Plugging these values and our boundary values for  into the arc length equation, we get:

\(\displaystyle L=\int_{0}^{4}\sqrt{(1)^{2}+(2)^{2}}dt\)

\(\displaystyle L=\int_{0}^{4}(\sqrt{1+4})dt\)

\(\displaystyle L=\int_{0}^{4}(\sqrt{5})dt\)

Now, using the Power Rule for Integrals

 for all ,

we can determine that:

\(\displaystyle L=[t\sqrt{5}]_{0}^{4}\textrm{}dt\)

\(\displaystyle L=(4)\sqrt{5}-(0)\sqrt{5}\)

\(\displaystyle L=4\sqrt{5}\)

In order to find the arc length, we must use the arc length formula for parametric curves:

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

Given \(\displaystyle x=4t-5\) and \(\displaystyle y=6t+2\), we can use using the Power Rule

for all  , to derive

\(\displaystyle \frac{dx}{dt}=(1)4t^{1-1}-(0)5=4\) and

\(\displaystyle \frac{dy}{dt}=(1)6t^{1-1}+(0)2=6\) .

Plugging these values and our boundary values for into the arc length equation, we get:

\(\displaystyle L=\int_{0}^{2}(\sqrt{(4)^{2}+(6)^{2}}dt\)

\(\displaystyle L=\int_{0}^{2}(\sqrt{16+36})dt\)

\(\displaystyle L=\int_{0}^{2}(\sqrt{52})dt\)

\(\displaystyle L=\int_{0}^{2}(\sqrt{4\times13})dt\)

\(\displaystyle L=\int_{0}^{2}(2\sqrt{13})dt\)

Now, using the Power Rule for Integrals

for all ,

we can determine that:

\(\displaystyle L=[2t\sqrt{13}]_{0}^{2}\textrm{}\)

\(\displaystyle L=2(2)\sqrt{13}-2(0)\sqrt{13}\)

\(\displaystyle L=4\sqrt{13}\)

In order to find the arc length, we must use the arc length formula for parametric curves:

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

Given \(\displaystyle x=9t+4\) and \(\displaystyle y=4t-1\), we can use using the Power Rule

 for all , to derive 

\(\displaystyle \frac{dx}{dt}=(1)9t^{1-1}+(0)4=9\) and 

\(\displaystyle \frac{dy}{dt}=(1)4t^{1-1}-(0)1=4\).

Plugging these values and our boundary values for  into the arc length equation, we get:

\(\displaystyle L=\int_{0}^{3}(\sqrt{(9)^{2}+(4)^{2}}dt\)

\(\displaystyle L=\int_{0}^{3}(\sqrt{81+16})dt\)

\(\displaystyle L=\int_{0}^{3}(\sqrt{97})dt\)

Now, using the Power Rule for Integrals

 for all ,

we can determine that:

\(\displaystyle L=[t\sqrt{97}]_{0}^{3}\textrm{}\)

\(\displaystyle L=(3)\sqrt{97}-(0)\sqrt{97}\)

\(\displaystyle L=3\sqrt{97}\)

In order to find the arc length, we must use the arc length formula for parametric curves:

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

Given \(\displaystyle x=t-11\) and \(\displaystyle y=7t+6\), we can use using the Power Rule

 for all , to derive 

\(\displaystyle \frac{dx}{dt}=(1)t^{1-1}-(0)11=1\)  and 

\(\displaystyle \frac{dy}{dt}=(1)7t^{1-1}+(0)6=7\).

Plugging these values and our boundary values for  into the arc length equation, we get:

\(\displaystyle L=\int_{0}^{4}(\sqrt{(1)^{2}+(7)^{2}}dt\)

\(\displaystyle L=\int_{0}^{4}(\sqrt{1+49})dt\)

\(\displaystyle L=\int_{0}^{4}(\sqrt{50})dt\)

\(\displaystyle L=\int_{0}^{4}(\sqrt{25\times2})dt\)

\(\displaystyle L=\int_{0}^{4}(5\sqrt{2})dt\)

Now, using the Power Rule for Integrals

 for all ,

we can determine that:

\(\displaystyle L=[5t\sqrt{2}]_{0}^{4}\textrm{}\)

\(\displaystyle L=5(4)\sqrt{2}-5(0)\sqrt{2}\)

\(\displaystyle L=20\sqrt{2}\)

To eliminate the parameter \(\displaystyle t\) from \(\displaystyle x\) and \(\displaystyle y\), we will solve the \(\displaystyle y\) equation for \(\displaystyle t\) and substitute the new expression into the \(\displaystyle x\) equation. We could also solve the \(\displaystyle x\) equation for \(\displaystyle t\) and substitute the new expression into the \(\displaystyle y\) equation, depending on which is easier.

For our equations, \(\displaystyle x=t^{3}+3t^{2}-3t-12\) and \(\displaystyle y=t-8\), it is easiest to solve the \(\displaystyle y\) equation for \(\displaystyle t\), giving us \(\displaystyle t=y+8\).  

Substituting our new expression for \(\displaystyle t\) into the \(\displaystyle x\) equation, we get

\(\displaystyle x=(y+8)^{3}+3(y+8)^{2}-3(y+8)-12\)

To eliminate the parameter \(\displaystyle t\) from \(\displaystyle x\) and \(\displaystyle y\), we will solve the \(\displaystyle y\) equation for \(\displaystyle t\) and substitute the new expression into the \(\displaystyle x\) equation. We could also solve the \(\displaystyle x\) equation for \(\displaystyle t\) and substitute the new expression into the \(\displaystyle y\) equation, depending on which is easier. 

For our equations,  \(\displaystyle x=ln(t)\) and \(\displaystyle y=ln(4t)\) , we will rearrange the \(\displaystyle y\) equation

\(\displaystyle y=ln(4t)\)

To eleiminate the \(\displaystyle ln\) on the right side of the equation, we will take the exponential of both sides of the equation

\(\displaystyle e^{y}=e^{ln(4t)}\)

Using the exponential identity \(\displaystyle e^{ln(x)}=x\)

\(\displaystyle e^{y}=4t\)

\(\displaystyle \frac{1}{4}e^{y}=t\)

Substituting this value of \(\displaystyle t\) into the \(\displaystyle x\) equation, we have

\(\displaystyle x=ln(\frac{1}{4}e^{y})\)

Using the logarithmic identity, \(\displaystyle ln(xy)=ln(x)+ln(y)\)

\(\displaystyle ln(\frac{1}{4}e^{y})=ln(\frac{1}{4})+ln(e^{y})\)

The using the identity, \(\displaystyle ln(e^{x})=x\)

\(\displaystyle ln(\frac{1}{4})+ln(e^{y})=ln(\frac{1}{4})+y\)

Giving us the final expression

\(\displaystyle x=ln(\frac{1}{4})+y\)

To eliminate the parameter \(\displaystyle t\) from \(\displaystyle x\) and \(\displaystyle y\), we will solve the \(\displaystyle y\) equation for \(\displaystyle t\) and substitute the new expression into the \(\displaystyle x\) equation. We could also solve the \(\displaystyle x\) equation for \(\displaystyle t\) and substitute the new expression into the \(\displaystyle y\) equation, depending on which is easier. 

For our equations, \(\displaystyle x=e^{2t}\) and \(\displaystyle y=e^{5t}\), we will rearrange the \(\displaystyle x\) equation.

\(\displaystyle x=e^{2t}\)

To eliminate the exponential from the right side of the equation, we will take the \(\displaystyle ln\) of both sides of the equation.

\(\displaystyle ln(x)=ln(e^{2t})\)

Using the logarithmic identity, \(\displaystyle ln(e^{x})=x\)

\(\displaystyle ln(x)=2t\)

\(\displaystyle \frac{1}{2}ln(x)=t\)

Substituting this value of \(\displaystyle t\) into the \(\displaystyle y\) equation, we have

\(\displaystyle y=e^{5*\frac{1}{2}ln(x)}=e^{10ln(x)}\)

Using the logarithmic identity \(\displaystyle cln(x)=ln(x^{c})\), where \(\displaystyle c\) is a constant

\(\displaystyle e^{10ln(x)}=e^{ln(x^{10})}=x^{10}\)

Therefore \(\displaystyle y=x^{10}\).