Add. Maths (checking of answers)

Audi SG

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.

deepak.c SG

1) You forgot square root of 54

2) It's correct

3) I going out for lunch liao, so never check.

Audi SG

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

deepak.c SG

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.

deepak.c SG

What do you mean by "eight h cm such that the height is thrice the radius"?

I am eating while I am typing this.

Forbiddensinner SG

Some doubts regarding the following questions.

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.

Forbiddensinner SG

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.

deepak.c SG

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

Audi SG

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

deepak.c SG

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

deepak.c SG

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

Audi SG

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?

deepak.c SG

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.

Audi SG

Typo for first sentence, I corrected already.

I am not perfect, else I would be god.

No worries. Thanks for helping. (:

Audi SG

Hey, but then again, what makes my method invalid? I don't think there's any mistake made in my method? ;p

deepak.c SG

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.