Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #41 : Parametric, Polar, And Vector

Given \displaystyle x=3t+7 and \displaystyle y=7t+3, what is \displaystyle y in terms \displaystyle x (rectangular form)?

Possible Answers:

\displaystyle y=\frac{7}{3}(x+7)+3

\displaystyle y=\frac{3}{7}(x+7)+3

\displaystyle y=\frac{7}{3}(x-7)-3

\displaystyle y=\frac{7}{3}(x-7)+3

\displaystyle y=\frac{3}{7}(x-7)+3

Correct answer:

\displaystyle y=\frac{7}{3}(x-7)+3

Explanation:

Given \displaystyle x=3t+7 and \displaystyle y=7t+3, let's solve both equations for \displaystyle t:

\displaystyle x=3t+7\rightarrow t=\frac{x-7}{3}

\displaystyle y=7t+3\rightarrow t=\frac{y-3}{7}

Since both equations equal \displaystyle t, let's set them equal to each other and solve for \displaystyle y:

\displaystyle \frac{y-3}{7}=\frac{x-7}{3}

\displaystyle y-3=7(\frac{x-7}{3})

\displaystyle y=\frac{7}{3}(x-7)+3

 

Example Question #42 : Parametric, Polar, And Vector

Given \displaystyle x=7t+5 and \displaystyle y=11t+2, what is \displaystyle y in terms of \displaystyle x (rectangular form)?

Possible Answers:

\displaystyle y=\frac{11}{7}(x-5)-2

\displaystyle y=\frac{11}{7}(x+5)-2

\displaystyle y=\frac{11}{7}(x-5)+2

None of the above

\displaystyle y=\frac{11}{7}(x+5)+2

Correct answer:

\displaystyle y=\frac{11}{7}(x-5)+2

Explanation:

Given \displaystyle x=7t+5 and \displaystyle y=11t+2, let's solve both equations for \displaystyle t:

\displaystyle x=7t+5\rightarrow t=\frac{x-5}{7}

\displaystyle y=11t+2\rightarrow t=\frac{y-2}{11}

Since both equations equal \displaystyle t, let's set them equal to each other and solve for \displaystyle y:

\displaystyle \frac{y-2}{11}=\frac{x-5}{7}

\displaystyle y-2=11(\frac{x-5}{7})

\displaystyle y=\frac{11}{7}(x-5)+2

 

Example Question #43 : Parametric, Polar, And Vector

Given \displaystyle x=-6t+11 and \displaystyle y=2t-10, what is \displaystyle y in terms of \displaystyle x (rectangular form)?

Possible Answers:

\displaystyle y=\frac{x-11}{3}-10

None of the above

\displaystyle y=-\frac{x-11}{3}-10

\displaystyle y=-\frac{x-11}{3}+10

\displaystyle y=\frac{x-11}{3}+10

Correct answer:

\displaystyle y=-\frac{x-11}{3}-10

Explanation:

Given \displaystyle x=-6t+11 and \displaystyle y=2t-10 let's solve both equations for \displaystyle t:

\displaystyle x=-6t+11\rightarrow t=-\frac{x-11}{6}

\displaystyle y=2t-10\rightarrow t=\frac{y+10}{2}

Since both equations equal \displaystyle t, let's set them equal to each other and solve for \displaystyle y:

\displaystyle \frac{y+10}{2}=-\frac{x-11}{6}

\displaystyle y+10=-2(\frac{x-11}{6})

\displaystyle y=-\frac{x-11}{3}-10

Example Question #44 : Parametric, Polar, And Vector

Convert the following equation from parametric to rectangular coordinates:

\displaystyle x=2t^2, y=10t

Possible Answers:

\displaystyle y=-10\sqrt{\frac{x}{2}}

\displaystyle y=\pm10\sqrt{\frac{x}{2}}

\displaystyle y=10\sqrt{\frac{x}{2}}

\displaystyle y=5x

Correct answer:

\displaystyle y=\pm10\sqrt{\frac{x}{2}}

Explanation:

To convert the equation from parametric form to rectangular form, we must eliminate the parameter. To do this, we can solve for t with respect to x:

\displaystyle t=\pm \sqrt{\frac{x}{2}}

The plus or minus is important to remember because a square root was taken.

Now, simply plug this into the equation for y:

\displaystyle y=\pm10\sqrt{\frac{x}{2}}

Example Question #45 : Parametric, Polar, And Vector

Convert the following to rectangular form from parametric form:

\displaystyle x=2t^3+5\displaystyle y=t

Possible Answers:

\displaystyle y=\left(\frac{x-5}{2}\right)^{\frac{3}{2}}

\displaystyle y=\sqrt[3]{\left(\frac{x-5}{2}\right)}

\displaystyle y=\frac{x-5}{2}

\displaystyle y=x

Correct answer:

\displaystyle y=\sqrt[3]{\left(\frac{x-5}{2}\right)}

Explanation:

To convert a parametric equation into a rectangular equation, we must eliminate the parameter. We were already given an equation where t was in terms of just a variable:

\displaystyle t=y

Next, substitute this into the equation for x which contains t:

\displaystyle x=2(y^3)+5

Finally, rearrage and solve for y:

\displaystyle y=\sqrt[3]{\left(\frac{x-5}{2}\right)}

Example Question #41 : Parametric, Polar, And Vector

Convert the following equation into rectanglar form:

\displaystyle x=\cos(t), y=cos^2 (t)+3

 

Possible Answers:

\displaystyle y=x^2+3

\displaystyle y=\cos(x)+3

\displaystyle y=-x^2+3

\displaystyle y=\arccos(x)+3

Correct answer:

\displaystyle y=x^2+3

Explanation:

To convert the given parametric equation into rectangualr coordinates, we must eliminate the parameter by solving for t:

\displaystyle t=\arccos (x)

Now replace t with the new term in the equation for y:

\displaystyle y= (\cos(\textup{arccos}(x)^2)+3=x^2+3

Example Question #41 : Parametric Form

Convert the following equation from parametric to rectangular form:

\displaystyle x=e^t, y=2\sin(t)

Possible Answers:

\displaystyle y=2\sin(\ln(x))

\displaystyle y=2\sin(e^t)

\displaystyle y=-2\sin(\ln(x))

\displaystyle y=x

Correct answer:

\displaystyle y=2\sin(\ln(x))

Explanation:

To convert from parametric to rectangular form, eliminate the parameter (t) from one of the equations:

\displaystyle t=\ln(x)

Now plug this into the equation for y to get our final answer:

\displaystyle y=2\sin(\ln(x))

Example Question #48 : Parametric, Polar, And Vector

Given \displaystyle x=3t-6 and \displaystyle y=7t+9, what is \displaystyle y in terms of \displaystyle x (rectangular form)?

Possible Answers:

\displaystyle y=\frac{7}{3}x+23

\displaystyle y=\frac{7}{3}x-23

\displaystyle y=23x-\frac{7}{3}

\displaystyle y=23x+\frac{7}{3}

None of the above

Correct answer:

\displaystyle y=\frac{7}{3}x+23

Explanation:

Given \displaystyle x=3t-6 and \displaystyle y=7t+9, let's solve both equations for \displaystyle t:

\displaystyle x=3t-6\rightarrow t=\frac{x+6}{3}

\displaystyle y=7t+9\rightarrow t=\frac{y-9}{7}

Since both equations equal \displaystyle t, let's set them equal to each other and solve for \displaystyle y:

\displaystyle \frac{y-9}{7}=\frac{x+6}{3}

\displaystyle y-9=\frac{7}{3}(x+6)

\displaystyle y-9=\frac{7}{3}x+14

\displaystyle y=\frac{7}{3}x+23

Example Question #49 : Parametric, Polar, And Vector

Given \displaystyle x=2t+11 and \displaystyle y=11t+2, what is \displaystyle y in terms of \displaystyle x (rectangular form)?

Possible Answers:

\displaystyle y=\frac{11}{2}(x-11)+2

\displaystyle y=\frac{11}{2}(x+11)+2

\displaystyle y=\frac{11}{2}(x-11)-2

None of the above

\displaystyle y=\frac{11}{2}(x+11)-2

Correct answer:

\displaystyle y=\frac{11}{2}(x-11)+2

Explanation:

Given \displaystyle x=2t+11 and \displaystyle y=11t+2,  let's solve both equations for \displaystyle t:

\displaystyle x=2t+11\rightarrow t=\frac{x-11}{2}

\displaystyle y=11t+2\rightarrow t=\frac{y-2}{11}

Since both equations equal \displaystyle t, let's set them equal to each other and solve for \displaystyle y:

\displaystyle \frac{y-2}{11}=\frac{x-11}{2}

\displaystyle y-2=\frac{11}{2}(x-11)

\displaystyle y=\frac{11}{2}(x-11)+2

Example Question #50 : Parametric, Polar, And Vector

Given \displaystyle x=t+3 and \displaystyle y=\frac{1}{2}t-3, what is \displaystyle y in terms of \displaystyle x (rectangular form)?

Possible Answers:

None of the above

\displaystyle y=\frac{x+3}{2}-3

\displaystyle y=\frac{x+3}{2}+3

\displaystyle y=\frac{x-3}{2}-3

\displaystyle y=\frac{x-3}{2}+3

Correct answer:

\displaystyle y=\frac{x-3}{2}-3

Explanation:

Given \displaystyle x=t+3 and \displaystyle y=\frac{1}{2}t-3,  let's solve both equations for \displaystyle t:

\displaystyle x=t+3\rightarrow t=x-3

\displaystyle y=\frac{1}{2}t-3\rightarrow t=2(y+3)

Since both equations equal \displaystyle t, let's set them equal to each other and solve for \displaystyle y:

\displaystyle 2(y+3)=x-3

\displaystyle y+3=\frac{x-3}{2}

\displaystyle y=\frac{x-3}{2}-3

 

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