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Question
Both the radius and the height of a circular cone change at the rate of 2 cm/s. How fast is the volume of the cone increasing when the radius= 10 cm and the height= 20 cm.
The volume of the con is : (1/3)*pi*r^2*h. Weknow that :
h(t)=2t & r(t)=2t. So, v(t)=(1/3)*pi*(4t^2)*2t=(1/3)*pi*8t^3.
Now we have to find the rate of change in the volume , so we
derive v(t) :
v'(t)=8*pi*t^2.
1) When the radius is 10 -> 10=2t1 -> t1=5 sec, so
v'(5)=8*pi*25
2) When the height is 20 -> 20=2t2 -> t2=10 sec, so
v'(10)=8*pi*100
Alon.
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Answers by Expert:
Alon Mandes
Expertise
Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
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1. I'm a team member of mathnerds (math site for answering questions)
2. I'm a team member in the Student's Union of the Technion, helping
students who have problems in mathematics.
3. 2 years of experience as a math teacher in college.
4. I give free homework help for high school students in
Mathematics & Physics.
5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" ,
"Complex Functions".
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Education/Credentials
M.A in Mathematics & Bs.c in Electronics.
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