C4 Rate of change Watch

bigmansouf
Badges: 20
Rep:
?
#1
Report Thread starter 4 weeks ago
#1
Q:
A hollow right circular cone is held vertex downwards beneath a tap leaking at the  2 cm^3/s. Find the rate of rise of water level when depth is 6cm given that the height of the cone is 18cm and radius is 12cm.

My attempt:

Name:  a1.jpg
Views: 5
Size:  25.1 KB

volume of cone =  \Pi r^{2}\frac{h}{3}
 \frac{\mathrm{d} v}{\mathrm{d} t}=2cm^{3}/s
r =12
thus

volume of cone =  \Pi (12)^{2}\frac{h}{3}
volume of cone = 48\Pih

 \frac{\mathrm{d} v}{\mathrm{d} h}= 48\Pi


 \frac{\mathrm{d} h}{\mathrm{d} t}= \frac{\mathrm{d} h}{\mathrm{d} v}\times \frac{\mathrm{d} v}{\mathrm{d} t}



 \frac{\mathrm{d} h}{\mathrm{d} t} = \frac{1}{48\Pi } \times 2 = \frac{1}{24\Pi } cm/s



however the answer the book gives is \frac{1}{8\Pi } cm/s

please help
0
reply
mqb2766
Badges: 12
Rep:
?
#2
Report 4 weeks ago
#2
(Original post by bigmansouf)
Q:
A hollow right circular cone is held vertex downwards beneath a tap leaking at the  2 cm^3/s. Find the rate of rise of water level when depth is 6cm given that the height of the cone is 18cm and radius is 12cm.

My attempt:

Name:  a1.jpg
Views: 5
Size:  25.1 KB

volume of cone =  \Pi r^{2}\frac{h}{3}
 \frac{\mathrm{d} v}{\mathrm{d} t}=2cm^{3}/s
r =12
thus

volume of cone =  \Pi (12)^{2}\frac{h}{3}
volume of cone = 48\Pih

 \frac{\mathrm{d} v}{\mathrm{d} h}= 48\Pi


 \frac{\mathrm{d} h}{\mathrm{d} t}= \frac{\mathrm{d} h}{\mathrm{d} v}\times \frac{\mathrm{d} v}{\mathrm{d} t}



 \frac{\mathrm{d} h}{\mathrm{d} t} = \frac{1}{48\Pi } \times 2 = \frac{1}{24\Pi } cm/s



however the answer the book gives is \frac{1}{8\Pi } cm/s

please help
I think you need to get the basic rate of change of the cone right. You know that the "base" radius is 2/3 of the height and you're evaluating when h = 6. So write the volume of the water (not the total cone) in terms of "h", differentiate it and sub it into the expression for dh/dt.
0
reply
bigmansouf
Badges: 20
Rep:
?
#3
Report Thread starter 3 weeks ago
#3
(Original post by mqb2766)
I think you need to get the basic rate of change of the cone right. You know that the "base" radius is 2/3 of the height and you're evaluating when h = 6. So write the volume of the water (not the total cone) in terms of "h", differentiate it and sub it into the expression for dh/dt.
thank you
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
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

  • Birkbeck, University of London
    Undergraduate Open Day - Startford Undergraduate
    Thu, 21 Mar '19
  • University of Wolverhampton
    Postgraduate Open Evening Postgraduate
    Thu, 21 Mar '19
  • Edge Hill University
    Undergraduate and Postgraduate - Campus Tour Undergraduate
    Fri, 22 Mar '19

Where do you need more help?

Which Uni should I go to? (68)
15.74%
How successful will I become if I take my planned subjects? (44)
10.19%
How happy will I be if I take this career? (81)
18.75%
How do I achieve my dream Uni placement? (61)
14.12%
What should I study to achieve my dream career? (46)
10.65%
How can I be the best version of myself? (132)
30.56%

Watched Threads

View All