Rates of change? Question. (1 Viewer)

[Skeptyks]

Just starting this topic and I have two question:

1. A chute drops sand at a constant rate of 8m^3/min. As the sand falls to the ground it forms a cone shape such that the height of the cone is twice its radius. Find the rate at which the height of the sand will be changing, to 2 decimal places, when its height is 2m.

Are we supposed to find the height of the cone in terms of r and then differentiate followed by differentiating the rate of change and multiplying them together and then finally substituting r=2/h back in for when "the height is 2m"?

2. The surface area of a spherical bubble is increasing at a constant rate of 1.9mm^2/second. Fine the rate of increase in its volume when its radius is 0.6mm.

I differentiate surface area with respect to r, flip it over and multiply with the differential of the rate of change of the surface area. When I substitute r=0.6, I get 19/48pi. Now I differentiate V with respect to A, multiply it by 19/48pi and in the end I get 0.0377 which is the wrong answer.

Rather simple questions probably but I don't really know what to do...

Any help is appreciated, thanks.

 

[Alkenes]

1) sub r=h/2 in the volume formula for cone, differentitate and then you know what to do ...

2) find dr/dt by using the rate of suface area and then use that to find volume rate

Just starting this topic and I have two question:

1. A chute drops sand at a constant rate of 8m^3/min. As the sand falls to the ground it forms a cone shape such that the height of the cone is twice its radius. Find the rate at which the height of the sand will be changing, to 2 decimal places, when its height is 2m.

Are we supposed to find the height of the cone in terms of r and then differentiate followed by differentiating the rate of change and multiplying them together and then finally substituting r=2/h back in for when "the height is 2m"?

2. The surface area of a spherical bubble is increasing at a constant rate of 1.9mm^2/second. Fine the rate of increase in its volume when its radius is 0.6mm.

I differentiate surface area with respect to r, flip it over and multiply with the differential of the rate of change of the surface area. When I substitute r=0.6, I get 19/48pi. Now I differentiate V with respect to A, multiply it by 19/48pi and in the end I get 0.0377 which is the wrong answer.

Rather simple questions probably but I don't really know what to do...

Any help is appreciated, thanks.

 

[fortune_cookie]

1.

Volume of a cone is V= (1/3) pi r^2 h

We know that h=2r

Also, note we want to find something h is a certain value. Therefore, express V in terms of h only

r= (h/2)

So, V= (1/3) pi [h/2]^2 h = (1/12) pi h^3

We are given how volume changes with time (i.e. dV/dt)

We want to find how height changes with time (dh/dt)

Make a chain rule

dh/dt = dh/dV x dV/dt

Now, dV/dh = (1/4) pi h^2

Therefore,

dh/dt = 1/[ (1/4) pi h^2 ] x 8 = 32/[ pi h^2 ]

When h=2 , dh/dt = 32/[4 pi] = 8/pi

Therefore the height is increasing at 8/pi metres per minute.

 

[Alkenes]

2)

 

[Skeptyks]

1.

Volume of a cone is V= (1/3) pi r^2 h

We know that h=2r

Also, note we want to find something h is a certain value. Therefore, express V in terms of h only

r= (h/2)

So, V= (1/3) pi [h/2]^2 h = (1/12) pi h^3

We are given how volume changes with time (i.e. dV/dt)

We want to find how height changes with time (dh/dt)

Make a chain rule

dh/dt = dh/dV x dV/dt

Now, dV/dh = (1/4) pi h^2

Therefore,

dh/dt = 1/[ (1/4) pi h^2 ] x 8 = 32/[ pi h^2 ]

When h=2 , dh/dt = 32/[4 pi] = 8/pi

Therefore the height is increasing at 8/pi metres per minute.

Answer is 2.55m per minute.

View attachment 25334

2)

Thanks heaps, I definitely did it the hardest way possible -.-;

 

[fortune_cookie]

2.

Given dA/dt = +1.9

We want how volume changes with time (dV/dt)

The formula for the area of a sphere is A= 4 pi r^2

For a chain rule using the formula of the object in question (area of sphere in this case) and the given information (dA/dt)

dV/dt = (dV/dA) x (dA/dt)

Now, we only have a formula in terms of r. You need to somehow introduce r into the equations

You can do this by spilting dV/dA into dV/dr x dr/dA

Volume of a sphere is V= (4/3) pi r^3 and Area of a Sphere (A) is above

dV/dr = 4 pi r^2 and dA/dr = 8 pi r

Therefore, dV/dt = [4 pi r^2 ] x 1/[8 pi r] x 1.9 = 0.95r

When r=0.6, dV/dt = 0.95 x 0.6 = 0.57mm^3 / second


Note the units, it's important you use the same units as in question.

 

[fortune_cookie]

Answer is 2.55m per minute.



Thanks heaps, I definitely did it the hardest way possible -.-;

Type 8/pi into calculator and see what you get.

 

[Skeptyks]

2.

Given dA/dt = +1.9

We want how volume changes with time (dV/dt)

The formula for the area of a sphere is A= 4 pi r^2

For a chain rule using the formula of the object in question (area of sphere in this case) and the given information (dA/dt)

dV/dt = (dV/dA) x (dA/dt)

Now, we only have a formula in terms of r. You need to somehow introduce r into the equations

You can do this by spilting dV/dA into dV/dr x dr/dA

Volume of a sphere is V= (4/3) pi r^3 and Area of a Sphere (A) is above

dV/dr = 4 pi r^2 and dA/dr = 8 pi r

Therefore, dV/dt = [4 pi r^2 ] x 1/[8 pi r] x 1.9 = 0.95r

When r=0.6, dV/dt = 0.95 x 0.6 = 0.57mm^3 / second


Note the units, it's important you use the same units as in question.

Thanks as well haha, I nearly forgot about the mm^3s^-1. Answer to question 1 is a bit different to answer though.

EDIT: MY BAD. I read your solutions wrong, sigh... I read "8pi" and I was like, this is definitely not 2.55, also thanks for the exact form tip. The answers round to 2 decimals without the question stating for it so I guess from now on I will just leave it in exact form.

Thanks everyone.

 

[fortune_cookie]

Thanks as well haha, I nearly forgot about the mm^3s^-1. Answer to question 1 is a bit different to answer though.

Yeah, I didn't read the question and give it as a decimal. Type 8/pi into a calcuator though and you will get 2.55.

 

[fortune_cookie]

Thanks as well haha, I nearly forgot about the mm^3s^-1. Answer to question 1 is a bit different to answer though.

EDIT: MY BAD. I read your solutions wrong, sigh... I read "8pi" and I was like, this is definitely not 2.55, also thanks for the exact form tip. The answers round to 2 decimals without the question stating for it so I guess from now on I will just leave it in exact form.
Thanks everyone.

The question does say "to two decimal places". If the question specifies decimal places then you have to give decimal form or you will lose a mark. If it doesn't say to give your answer as a decimal, then you can give as exact. Just make sure you read the question in the exam.