Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a)[PR1] How do you find the slope of a tangent line to a polar curve?

(b) How do you find the area of a region bounded by a polar curve?

(c) How do you find the length of a polar curve?

Short Answer

Expert verified
  1. For a polar curve the slope of the tangent is\({\rm{M = }}\frac{{\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}} \cdot {\rm{sin(\theta ) + r}} \cdot {\rm{cos(\theta )}}}}{{\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}} \cdot {\rm{cos(\theta ) - r}} \cdot {\rm{sin(\theta )}}}}\).
  2. For a polar curve the of a region is\({\rm{A = }}\int_{\rm{\alpha }}^{\rm{\beta }} {\frac{{\rm{1}}}{{\rm{2}}} \cdot } {{\rm{r}}^{\rm{2}}}{\rm{d\theta }}\).
  3. For a polar curve the expression for its length is \({\rm{L = }}\int_{\rm{a}}^{\rm{b}} {\sqrt {{{\rm{r}}^{\rm{2}}}{\rm{ + }}{{\left( {\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}}} \right)}^{\rm{2}}}} {\rm{d\theta }}} \).

Step by step solution

01

Concept Introduction

Polar curves are graphs of equations with polar coordinates as their basis. The polar curve, like ordinary equations and curves, is made up of all polar coordinates that satisfy the specified equation.

02

Slope of tangent to Polar Curve

(a)

Mathematically polar curve is defined as –

\(\begin{aligned}{c}{\rm{f(x,y) = f(r,\theta )}}\\{\rm{ = rcos(\theta ) + rsin(\theta )}}\end{aligned}\)

Now tangent is basically –

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}{\rm{ = }}\frac{{{{{\rm{dy}}} \mathord{\left/

{\vphantom {{{\rm{dy}}} {{\rm{d\theta }}}}} \right.

\kern-\nulldelimiterspace} {{\rm{d\theta }}}}}}{{{{{\rm{dx}}} \mathord{\left/

{\vphantom {{{\rm{dx}}} {{\rm{d\theta }}}}} \right.

\kern-\nulldelimiterspace} {{\rm{d\theta }}}}}}\)

Hence, the slope of tangent\({\rm{M}}\)will be –

\({\rm{M = }}\frac{{\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}} \cdot {\rm{sin(\theta ) + r}} \cdot {\rm{cos(\theta )}}}}{{\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}} \cdot {\rm{cos(\theta ) - r}} \cdot {\rm{sin(\theta )}}}}\)

Therefore, the result is obtained as \({\rm{M = }}\frac{{\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}} \cdot {\rm{sin(\theta ) + r}} \cdot {\rm{cos(\theta )}}}}{{\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}} \cdot {\rm{cos(\theta ) - r}} \cdot {\rm{sin(\theta )}}}}\).

03

Area of region bounded by Polar Curve

(b)

Area of polar coordinates (say\({\rm{A}}\)) is calculated by standard formula –

\({\rm{A = }}\int_{\rm{\alpha }}^{\rm{\beta }} {\frac{{\rm{1}}}{{\rm{2}}} \cdot } {{\rm{r}}^{\rm{2}}}{\rm{d\theta }}\)

This gives the area enclosed by below figure –

Therefore, the result is obtained as \({\rm{A = }}\int_{\rm{\alpha }}^{\rm{\beta }} {\frac{{\rm{1}}}{{\rm{2}}} \cdot } {{\rm{r}}^{\rm{2}}}{\rm{d\theta }}\).

04

Length of a Polar Curve

(c)

Mathematically polar curve is defined as –

\({\rm{r = f(\theta ), \theta }} \in \left( {{\rm{a,b}}} \right)\)

Then the length of polar curve (say \({\rm{L}}\)) is –

\({\rm{L = }}\int_{\rm{a}}^{\rm{b}} {\sqrt {{{\rm{r}}^{\rm{2}}}{\rm{ + }}{{\left( {\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}}} \right)}^{\rm{2}}}} {\rm{d\theta }}} \)

Therefore, the result is obtained as \({\rm{L = }}\int_{\rm{a}}^{\rm{b}} {\sqrt {{{\rm{r}}^{\rm{2}}}{\rm{ + }}{{\left( {\frac{{{\rm{dr}}}}{{{\rm{d\theta }}}}} \right)}^{\rm{2}}}} {\rm{d\theta }}} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free