Given equations for \(\displaystyle y\) and \(\displaystyle x\) in terms of \(\displaystyle t\), we can find the derivative of parametric equations as follows:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), as the \(\displaystyle dt\) terms will cancel out.

Using the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) and given \(\displaystyle y=7t^{2}-12\) and \(\displaystyle x=5t^{2}+2\):

\(\displaystyle \frac{dy}{dt}=(2)7t^{2-1}-(0)12=14t\)

\(\displaystyle \frac{dx}{dt}=(2)5t^{2-1}+(0)2=10t\)

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{14t}{10t}=\frac{7}{5}\).

Given equations for \(\displaystyle y\) and \(\displaystyle x\) in terms of \(\displaystyle t\), we can find the derivative of parametric equations as follows:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), as the \(\displaystyle dt\) terms will cancel out.

Using the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) and given \(\displaystyle y=9t^{2}-13\) and \(\displaystyle x=t^{2}+15\):

\(\displaystyle \frac{dy}{dt}=(2)9t^{2-1}-(0)13=18t\)

\(\displaystyle \frac{dx}{dt}=(2)t^{2-1}+15=2t\)

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{18t}{2t}=9\)

Since we have two equations \(\displaystyle y=2t-9\) and \(\displaystyle x=5t+10\), we can find \(\displaystyle \frac{dy}{dx}\) by dividing the derivatives of the two equations - thus:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) since the \(\displaystyle dt\) terms cancel out by standard rules of division of fractions. 

In order to find the derivatives of \(\displaystyle y\) and \(\displaystyle x\), let's use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\):

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

\(\displaystyle x=5t+10\rightarrow\frac{dx}{dt}=(1)5t^{1-1}+(0)10=5\)

Therefore, 

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{5}\).

Since we have two equations \(\displaystyle y=7t+4\) and \(\displaystyle x=7t-8\), we can find \(\displaystyle \frac{dy}{dx}\) by dividing the derivatives of the two equations - thus:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) since the \(\displaystyle dt\) terms cancel out by standard rules of division of fractions. 

In order to find the derivatives of \(\displaystyle y\) and \(\displaystyle x\), let's use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\):

\(\displaystyle y=7t+4\rightarrow \frac{dy}{dt}=(1)7t^{1-1}+(0)4=7\)

\(\displaystyle x=7t-8\rightarrow\frac{dx}{dt}=(1)7t^{1-1}-(0)8=7\)

Therefore,

 \(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{7}{7}=1\).

Since we have two equations \(\displaystyle y=5t^{2}+6t-9\)  and \(\displaystyle x=2t^{2}-6t+12\), we can find \(\displaystyle \frac{dy}{dx}\) by dividing the derivatives of the two equations - thus:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) (since the \(\displaystyle dt\) terms cancel out by standard rules of division of fractions). 

In order to find the derivatives of \(\displaystyle y\) and \(\displaystyle x\), let's use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\):

\(\displaystyle y=5t^{2}+6t-9\rightarrow \frac{dy}{dt}=(2)5t^{2-1}+(1)6t^{1-1}-(0)9=10t+6\)

\(\displaystyle x=2t^{2}-6t-12\rightarrow \frac{dx}{dt}=(2)2t^{2-1}-(1)6t^{1-1}-(0)12=4t-6\)

Therefore, \(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{10t+6}{4t-6}=\frac{2(5t+3)}{2(2t-3)}=\frac{5t+3}{2t-3}\).

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  \(\displaystyle x=5t-1\) and \(\displaystyle y=2t+9\), we can use using the Power Rule

 for all  , to derive

\(\displaystyle \frac{dx}{dt}=(1)5t^{1-1}-(0)1=5\)  and \(\displaystyle \frac{dy}{dt}=(1)2t^{1-1}+(0)9=2\).

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

\(\displaystyle L=\int_{0}^{3}\sqrt{(5)^{2}+(2)^{2}}dt\)

\(\displaystyle L=\int_{0}^{3}(\sqrt{25+4})dt\)

\(\displaystyle L=\int_{0}^{3}(\sqrt{29})dt\)

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:

\(\displaystyle L=(t\sqrt{29})_{0}^{3}\textrm{}\)

\(\displaystyle L=3\sqrt{29}-0\sqrt{29}\)

\(\displaystyle L=3\sqrt{29}\)

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:

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

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

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