Add. Maths (checking of answers)
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Some doubts regarding the following questions.
Consider a cone of base radius r cm, eight h cm such that the height is thrice the radius. The radius of the cone is increasing at a rate of 5 cm/s. Find the rate of change of volume of the cone when the height is sq.rt 54 leaving your answer in terms of pi.
60 pi is the answer I got. My answer differs from my friends'. Will explain how I got my answer.
Firstly, dv/dt = dv/dr x dr/dt
To find volume, v = 1/3(PI)(SQ.RT 54)(r)^2
dv/dr = 2/3(PI)(SQ.RT 54)(r)
Since height is thrice of radius, we know that r = (SQ.RT 54) / 3
So dv/dr = 2/3(PI)(SQ.RT54)/3
dv/dt = 2/3(PI)(SQ.RT54)/3 X 5
= 60 pi cm3/s
Secondly,find dy/dx for y = tan (ln x )
My answer... sec^2(ln x)(1/x) ?
Lastly, find dy/dx = ln(sin^3 x)
Ans: 3 cot x
That's about all that I have to ask. Thank you.
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1) You forgot square root of 54
2) It's correct
3) I going out for lunch liao, so never check.
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Originally posted by deepak.c:
1) You forgot square root of 54
Which part did I leave out sq.rt 54?And anyway, one more question. Find the values of p and q if y = px - q / (x-1)(x-4) has a turning point at A (2,-1) State whether y is a maximum and minimum point at A.
Found that p= 1, q = 0. For d^2y/dx^2, I have 0.5 which means it is a minimum point?
Only have these questions that I am not that clear in. thanks
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Originally posted by Audi:
Which part did I leave out sq.rt 54?And anyway, one more question. Find the values of p and q if y = px - q / (x-1)(x-4) has a turning point at A (2,-1) State whether y is a maximum and minimum point at A.
Found that p= 1, q = 0. For d^2y/dx^2, I have 0.5 which means it is a minimum point?
Only have these questions that I am not that clear in. thanks
Think I better look at it later tonight, rushing out and I am late.
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What do you mean by "eight h cm such that the height is thrice the radius"?
I am eating while I am typing this.
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Originally posted by Audi:
Some doubts regarding the following questions.
Consider a cone of base radius r cm, eight h cm such that the height is thrice the radius. The radius of the cone is increasing at a rate of 5 cm/s. Find the rate of change of volume of the cone when the height is sq.rt 54 leaving your answer in terms of pi.
60 pi is the answer I got. My answer differs from my friends'. Will explain how I got my answer.
Firstly, dv/dt = dv/dr x dr/dt
To find volume, v = 1/3(PI)(SQ.RT 54)(r)^2
dv/dr = 2/3(PI)(SQ.RT 54)(r)
Since height is thrice of radius, we know that r = (SQ.RT 54) / 3
So dv/dr = 2/3(PI)(SQ.RT54)/3
dv/dt = 2/3(PI)(SQ.RT54)/3 X 5
= 60 pi cm3/s
Secondly,find dy/dx for y = tan (ln x )
My answer... sec^2(ln x)(1/x) ?
Lastly, find dy/dx = ln(sin^3 x)
Ans: 3 cot x
That's about all that I have to ask. Thank you.
Bolded part should be 2/3(PI)(SQ.RT54)(SQ.RT54)/3
Part 3 is correct.
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Originally posted by Audi:
Which part did I leave out sq.rt 54?And anyway, one more question. Find the values of p and q if y = px - q / (x-1)(x-4) has a turning point at A (2,-1) State whether y is a maximum and minimum point at A.
Found that p= 1, q = 0. For d^2y/dx^2, I have 0.5 which means it is a minimum point?
Only have these questions that I am not that clear in. thanks
Assuming your answers are correct, it is a min. point.
If d^2y/dx^2 is positive, it is a min. point, if d^2y/dx^2 is negative, it is a max. point.
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Originally posted by Forbiddensinner:
Bolded part should be 2/3(PI)(SQ.RT54)(SQ.RT54)/3
Part 3 is correct.
Then Audi's part 1 should be correct then.
Audi shouldn't worry too much about getting different answers from classmates, there are many different correct answers to one calculus questions (many ways to skin a calc).
y = ln (sin x)^3
let sin x = u
y = 3 ln u
y' = 3 (u'/u)
y' = 3 (cos x / sin x)
y' = 3 cot x
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Argh, I think the answer of 60 pi for the question on the cone could be incorrect. My teacher said she adapted the question from an online site so she offered us 4 answers.
1.810 pi
2. 270 pi
3. 15(sqrt 6) pi
4. 90 pi
All of which look impossible to me. I have tried the question umpteen times but still got 60 pi.
Thanks for helping
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Originally posted by Audi:
Argh, I think the answer of 60 pi for the question on the cone could be incorrect. My teacher said she adapted the question from an online site so she offered us 4 answers.
1.810 pi
2. 270 pi
3. 15(sqrt 6) pi
4. 90 pi
All of which look impossible to me. I have tried the question umpteen times but still got 60 pi.
Thanks for helping
Sorry, should be 90 Pi
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v = (pi) (r)^2 (h) / 3
h = 3r
v = (pi) (r)^2 (3r) / 3
v = 3 (pi) (r)^3 / 3
v = (pi) (r)^3
dv/dr = 3 (pi) (r)^2
dr/dt = 5 cm/s
when h = sqrt (54) ===>> r = sqrt (54) / 3
dv/dt = 3 (pi) (sqrt 54/3)^2 * 5
dv/dt = 90 pi
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Originally posted by deepak.c:
v = (pi) (r)^2 (h)
h = 3r
v = (pi) (r)^2 (3r) / 3
v = 3 (pi) (r)^3 / 3
v = (pi) (r)^3
dv/dr = 3 (pi) (r)^2
dr/dt = 5 cm/s
when h = sqrt (54) ===>> r = sqrt (54) / 3
dv/dt = 3 (pi) (sqrt 54/3)^2 * 5
dv/dt = 90 pi
why is the formula for a cylinder used in this case? -
Originally posted by Audi:
why is the formula for a cylinder used in this case?Typo for first sentence, I corrected already.
I am not perfect, else I would be god.
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Originally posted by deepak.c:
Typo for first sentence, I corrected already.
I am not perfect, else I would be god.
No worries. Thanks for helping. (: -
Hey, but then again, what makes my method invalid? I don't think there's any mistake made in my method? ;p
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Originally posted by Audi:
Hey, but then again, what makes my method invalid? I don't think there's any mistake made in my method? ;p
The cone height is related to the radius, you need to factor it into the volume of cone equation first before you differentiate.