miadiz
Badges: 8
Rep:
?
#1
Report Thread starter 1 year ago
#1
Need help with this problem:


A graph shows part of the curve C with parametric equations

x = (t+1)^2, y= 1/2 t^3 +3, t>=-1

P is the point on the curve where t=2. The line S is the normal to C at P.

a.) Find an equation of S

The shaded region R is bounded by C, S, the z-axis and the line with equation X=1.

b.) Using integration and showing all your working, find the area of R.

I've done part a. In part b I think I know what to do just not sure how to write out the integral. I don't know if I need to use the two parametric equations (and multiply or add them??) or use some other equation.
0
reply
Kevin De Bruyne
Badges: 21
Rep:
?
#2
Report 1 year ago
#2
(Original post by miadiz)
Need help with this problem:


A graph shows part of the curve C with parametric equations

x = (t+1)^2, y= 1/2 t^3 +3, t>=-1

P is the point on the curve where t=2. The line S is the normal to C at P.

a.) Find an equation of S

The shaded region R is bounded by C, S, the z-axis and the line with equation X=1.

b.) Using integration and showing all your working, find the area of R.

I've done part a. In part b I think I know what to do just not sure how to write out the integral. I don't know if I need to use the two parametric equations (and multiply or add them??) or use some other equation.
This is in TSR help.

I have asked for it to be moved to Maths
0
reply
Chittesh14
Badges: 19
Rep:
?
#3
Report 1 year ago
#3
(Original post by miadiz)
Need help with this problem:


A graph shows part of the curve C with parametric equations

x = (t+1)^2, y= 1/2 t^3 +3, t>=-1

P is the point on the curve where t=2. The line S is the normal to C at P.

a.) Find an equation of S

The shaded region R is bounded by C, S, the z-axis and the line with equation X=1.

b.) Using integration and showing all your working, find the area of R.

I've done part a. In part b I think I know what to do just not sure how to write out the integral. I don't know if I need to use the two parametric equations (and multiply or add them??) or use some other equation.
I assume you mean the x-axis and not the z-axis.

You need to use the formula of the area of a curve:
\int^1_0 y\ dx = \int^1_0 y\ (dx/dt) dt

Parametric integration.

You will have to also work out the area under the normal. I haven't drawn the graph out so I can't visualise it, but it may involve subracting areas etc.

It'd be great if you can post a picture of the question and your working out !
0
reply
miadiz
Badges: 8
Rep:
?
#4
Report Thread starter 1 year ago
#4
(Original post by Chittesh14)
I assume you mean the x-axis and not the z-axis.

You need to use the formula of the area of a curve:
\int^1_0 y\ dx = \int^1_0 y\ (dx/dt) dt

Parametric integration.

You will have to also work out the area under the normal. I haven't drawn the graph out so I can't visualise it, but it may involve subracting areas etc.

It'd be great if you can post a picture of the question and your working out !
Yeah I don't know why I wasn't thinking of that formula... Thank you though

Name:  IMG_20180406_162447-compressed.jpg.jpeg
Views: 74
Size:  42.2 KB

Name:  IMG_20180406_162437-compressed.jpg.jpeg
Views: 73
Size:  42.4 KB
0
reply
Chittesh14
Badges: 19
Rep:
?
#5
Report 1 year ago
#5
(Original post by miadiz)
Yeah I don't know why I wasn't thinking of that formula... Thank you though

Name:  IMG_20180406_162447-compressed.jpg.jpeg
Views: 74
Size:  42.2 KB

Name:  IMG_20180406_162437-compressed.jpg.jpeg
Views: 73
Size:  42.4 KB
No worries .
Have you done it? If you haven't, I'll help you tomorrow, sorry for the late reply.
0
reply
miadiz
Badges: 8
Rep:
?
#6
Report Thread starter 1 year ago
#6
(Original post by Chittesh14)
No worries .
Have you done it? If you haven't, I'll help you tomorrow, sorry for the late reply.
Yeah, I did it. I got 58.9 as the area. By doing the integration with y and dx/dt ( = 2(t+1) ) and t limits 2 and 0. Then I added the previous value (34.4) to the area of the triangle (24.5).
0
reply
Chittesh14
Badges: 19
Rep:
?
#7
Report 1 year ago
#7
(Original post by miadiz)
Yeah, I did it. I got 58.9 as the area. By doing the integration with y and dx/dt ( = 2(t+1) ) and t limits 2 and 0. Then I added the previous value (34.4) to the area of the triangle (24.5).
Nice one . Post some more questions if you find anything else difficult!
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

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.

Personalise

University open days

  • Bournemouth University
    Undergraduate Open Day Undergraduate
    Wed, 19 Feb '20
  • Buckinghamshire New University
    Postgraduate and professional courses Postgraduate
    Wed, 19 Feb '20
  • University of Warwick
    Warwick Business School Postgraduate
    Thu, 20 Feb '20

Has your university offer been reduced?

Yes (19)
28.36%
No (36)
53.73%
Don't know (12)
17.91%

Watched Threads

View All