Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    9
    ReputationRep:
    I'm stuck on a question from the C4 book (chapter 6, Mixed Exercises 15 (d)).


    Here's the solution bank answer:


    I understand finding the area of the whole triangle, but not how to find the area of A1. Why is the curve 4sinθ(-5sinθ)? I know how to go from there, but not how they got that equation.

    Thanks guys!
    • TSR Support Team
    • Very Important Poster
    • Welcome Squad
    Offline

    20
    ReputationRep:
    TSR Support Team
    Very Important Poster
    Welcome Squad
    (Original post by кяя)
    I understand finding the area of the whole triangle, but not how to find the area of A1. Why is the curve 4sinθ(-5sinθ)? I know how to go from there, but not how they got that equation.

    Thanks guys!
    x=5 \cos \theta \Rightarrow dx=-5 \sin \theta \ \text{d} \theta

    \displaystyle \therefore \int y \ \text{d}x = \int 4 \sin \theta \times -5 \sin \theta \ \text{d} \theta
    Offline

    22
    ReputationRep:
    (Original post by кяя)
    ...
    As an alternative to the above (I prefer this once since you get used to the idea of substitutions being scalings and this will help once you do multivariate calculus and need to deal with Jacobians, but I digress):

    You know the area under a curve is given in cartesian co-ordinates as \int_{x=x_1}^{x = x_2} y \, \mathrm{d}x, yes? But at the same time you can think of \frac{\mathrm{d}x}{ \mathrm{d}\theta } \, \mathrm{d}\theta = \mathrm{d}x since they \mathrm{d}\theta's kind of cancel.

    Let's go ahead and re-write this now:

    \displaystyle

\begin{equation*} \int_{x = x_1}^{x = x_2} y \, \mathrm{d}x = \int_{\theta = \theta_1}^{\theta = \theta_2} y \, \frac{\mathrm{d}x}{ \mathrm{d}\theta } \, \mathrm{d}\theta\end{equation*}

    Now you know that x = 5 \cos \theta \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}\th  eta} = -5\sin \theta and \theta_1 = \frac{\pi}{4} and \theta_2 = 0 but then the negative sign means you flip them, so you're left with:

    \displaystyle 

\begin{equation*} \int_{\theta = 0}^{\theta = \frac{\pi}{4}} 4 \sin \theta \times -5\sin \theta \, \mathrm{d}\theta
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you rather give up salt or pepper?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.