Rates of Change (1 Viewer)

[Doctor Jolly]

The liquid in a vessel is allowed to flow out through a tap in the bottom at a rate given by dV/dt = -10(40 - t) where V cm^3 is the volume of liquid in the vessel at time t seconds.

Initially the vessel contained 8.0 liters of liquid. Find the amount of liquid in the vessel after 10 seconds.

answer: 4.5 L.

okay, where did I go wrong?

My guess is at my integration

Thanks.

 

[munch0r]

1 cm^3 = 1 milliliter

hence C = 8000 not 8
which gives you an answer of 4500mL which = 4.5L

 

[Doctor Jolly]

1 cm^3 = 1 milliliter

hence C = 8000 not 8
which gives you an answer of 4500mL which = 4.5L

Why change it to milliliter though?

 

[tommykins]

回复: Re: Rates of Change

your working out is to neat omfg.

Vcm^3 = 1 ml.

Inititally 8 L in vessel = 8000ml.

 

[Doctor Jolly]

Re: 回复: Re: Rates of Change

your working out is to neat omfg.

Vcm^3 = 1 ml.

Inititally 8 L in vessel = 8000ml.

haha thanks.

I think I get it why you change it to milliliter now

 

[tommykins]

回复: Re: 回复: Re: Rates of Change

The vessel is losing V cm^3/s which means its losing Vml/s.

 

[Pwnage101]

Re: 回复: Re: Rates of Change

your working out is to neat omfg.

agreed.

 

[Doctor Jolly]

Re: 回复: Re: Rates of Change

agreed.

haha. Thanks again.

 

[Doctor Jolly]

Sorry, I have another question

1. A boy is flying a kite. If the kite is 60m above the ground and the wind is blwoing it on a horizontal course at 8m/s, find how fast the boy is paying out the string when the kite is 100m from him.

answer: 6.4m/s

2.Two straight roads OX, OY intersect at right angles at O. A girl G starts from a point P, 21km from O on the road OX, and proceeds towards O at a speed of 3km/h. A boy B starts simultaneously from O and proceeds along OY at 4km/h. Find the rate at which the distance between them is changing after 2 hours.

answer: 13/17 km/h decreasing.

Thanks so much

 

[lyounamu]

Sorry, I have another question

1. A boy is flying a kite. If the kite is 60m above the ground and the wind is blwoing it on a horizontal course at 8m/s, find how fast the boy is paying out the string when the kite is 100m from him.

answer: 6.4m/s

2.Two straight roads OX, OY intersect at right angles at O. A girl G starts from a point P, 21km from O on the road OX, and proceeds towards O at a speed of 3km/h. A boy B starts simultaneously from O and proceeds along OY at 4km/h. Find the rate at which the distance between them is changing after 2 hours.

answer: 13/17 km/h decreasing.

Thanks so much

1.
Draw a right-angled triangle where hypotenuse is y, opposite is 60 and adjacent being x.

You are given dx/dt = 8 and you are trying to find dy/dt
dy/dt = dx/dt . dy/dx

And from the triangle you can deduce that
y^2 = x^2 + 60^2
Arrange that, you get:
x^2 = y^2-60^2
x = SQRT (y^2 - 60^2)
dx/dy = 1/2(y^2-60^2)^-1/2 . 2y
so dy/dx = reciprocal of what I wrote
When you substitute y=100, dy/dx = 4/5

So dy/dt = dx/dy . dy/dx = 8 . 4/5 = 6.4

2. Draw a triangle here too where hypotenuse = y, opposite = x and adjacent being z.

Hypotenuse is basically the distance between the two and the opposite is basically the girl's displacement with adjacent being the boy's displacement.

So x = 21 - 3t and z = 4t

Using y^2= x^2 + z^2
= (21-3t)^2 + (4t)^2 = 441 - 126t + 9t^2 + 16t^2
y = SQRT (441-126t + 25t^2)
dy/dt = 1/2 (-126 + 50t) (441-126t+25t^2)^-1/2

dy/dt = -13/17 when you sub t = 2 into dy/dt.

EDIT: Since we are trying to find the RATE at which the distance changes, you are trying to find dy/dt. And by finding the displacement in terms of t for y, I could find the derivative and sub t =2 to find the answer. My answer is dy/dt = -13/17 which means that the distance DECREASES by 13/17 per hour.

 

[Doctor Jolly]

1.
Draw a right-angled triangle where hypotenuse is y, opposite is 60 and adjacent being x.

You are given dx/dt = 8 and you are trying to find dy/dt
dy/dt = dx/dt . dy/dx

And from the triangle you can deduce that
y^2 = x^2 + 60^2
Arrange that, you get:
x^2 = y^2-60^2
x = SQRT (y^2 - 60^2)
dx/dy = 1/2(y^2-60^2)^-1/2 . 2y
so dy/dx = reciprocal of what I wrote
When you substitute y=100, dy/dx = 4/5

So dy/dt = dx/dy . dy/dx = 8 . 4/5 = 6.4

2. Draw a triangle here too where hypotenuse = y, opposite = x and adjacent being z.

Hypotenuse is basically the distance between the two and the opposite is basically the girl's displacement with adjacent being the boy's displacement.

So x = 21 - 3t and z = 4t

Using y^2= x^2 + z^2
= (21-3t)^2 + (4t)^2 = 441 - 126t + 9t^2 + 16t^2
y = SQRT (441-126t + 25t^2)
dy/dt = 1/2 (-126 + 50t) (441-126t+25t^2)^-1/2

dy/dt = -13/17 when you sub t = 2 into dy/dt.

EDIT: Since we are trying to find the RATE at which the distance changes, you are trying to find dy/dt. And by finding the displacement in terms of t for y, I could find the derivative and sub t =2 to find the answer. My answer is dy/dt = -13/17 which means that the distance DECREASES by 13/17 per hour.

THANKS SO MUCH Namu

I got it now I shall re-do them at this moment!

Did you perhaps mean "dy/dt = dx/dt . dy/dx = 8 . 4/5 = 6.4" instead of "dy/dt = dx/dy . dy/dx = 8 . 4/5 = 6.4" though?

 

[lyounamu]

With questions like above, you really need to draw a diagram. You need to be very visual person when you try to come about solving questions like that. They are extremely difficult.

AND one tip I can give you is that,

Before you try to solve the question, make sure you OUTLINE ALL THE NECESSARY INFORMATION such as dx/dt = 8 or whatever.

Then come about solving the question.

As in the first question I did, I actually didn't go straight into saying that

y^2 = x^2 + 60^2
so y = SQRT(x^2+60^2)
dy/dx = 1/2(x^2+60^2)^-1/2 . 2x

BECAUSE I cannot sub y=100 into this. THEREFORE, I went around a long way to find dy/dx by FLIPPING dx/dy.

So I actually found an entity for x first, differentiated it and flipped it so I have
a little nice equation like
dy/dx = 1/2(y^2-60^2)^-1/2 . 2y WHERE I CAN ACTUALLY SUB Y = 100.

Be wary of such a question.

For Question two, it was quite tedious. This type of question is quite rare. The format you follow is quite different from the first question.

Since you have too many variables with variable, t (time) coming into it, I strived to find an entity of y in terms of t.

But in the question, they actually told us about the each displacement component, x (vertical) and z (horizontal).

So I could actually find the distance (y) in terms of t. Because

y^2 = x^2 + z^2 (refer to the diagram that I initially suggested).

Then the rest should be quite easy to follow.

THIS IS NAMU'S BRIEF TUTORIAL ON THE TWO QUESTIONS ASKED. Hope my pathetic explanation didn't alienate you from the truth...

 

[lyounamu]

THANKS SO MUCH Namu

I got it now I shall re-do them at this moment!

Did you perhaps mean "dy/dt = dx/dt . dy/dx = 8 . 4/5 = 6.4" instead of "dy/dt = dx/dy . dy/dx = 8 . 4/5 = 6.4" though?

Yep, that's a typo. I actually never go back to what I write once I write (unless I know it's wrong from the start). So excuse my laziness.

 

[Doctor Jolly]

With questions like above, you really need to draw a diagram. You need to be very visual person when you try to come about solving questions like that. They are extremely difficult.

AND one tip I can give you is that,

Before you try to solve the question, make sure you OUTLINE ALL THE NECESSARY INFORMATION such as dx/dt = 8 or whatever.

Then come about solving the question.

As in the first question I did, I actually didn't go straight into saying that

y^2 = x^2 + 60^2
so y = SQRT(x^2+60^2)
dy/dx = 1/2(x^2+60^2)^-1/2 . 2x

BECAUSE I cannot sub y=100 into this. THEREFORE, I went around a long way to find dy/dx by FLIPPING dx/dy.

So I actually found an entity for x first, differentiated it and flipped it so I have
a little nice equation like
dy/dx = 1/2(y^2-60^2)^-1/2 . 2y WHERE I CAN ACTUALLY SUB Y = 100.

Be wary of such a question.

For Question two, it was quite tedious. This type of question is quite rare. The format you follow is quite different from the first question.

Since you have too many variables with variable, t (time) coming into it, I strived to find an entity of y in terms of t.

But in the question, they actually told us about the each displacement component, x (vertical) and z (horizontal).

So I could actually find the distance (y) in terms of t. Because

y^2 = x^2 + z^2 (refer to the diagram that I initially suggested).

Then the rest should be quite easy to follow.

THIS IS NAMU'S BRIEF TUTORIAL ON THE TWO QUESTIONS ASKED. Hope my pathetic explanation didn't alienate you from the truth...

haha.

Your tutorial was not at all brief. You handled the question exactly as how Cornoreos wanted it to be handled (as I'm using his textbook). It's just that I labelled my diagrams wrong and subbed y in a bit too early I just did about 5 of his questions and got them all right, so I'm very happy

Thanks again, Namu

 

[lolokay]

Sorry, I have another question

1. A boy is flying a kite. If the kite is 60m above the ground and the wind is blwoing it on a horizontal course at 8m/s, find how fast the boy is paying out the string when the kite is 100m from him.

answer: 6.4m/s

2.Two straight roads OX, OY intersect at right angles at O. A girl G starts from a point P, 21km from O on the road OX, and proceeds towards O at a speed of 3km/h. A boy B starts simultaneously from O and proceeds along OY at 4km/h. Find the rate at which the distance between them is changing after 2 hours.

answer: 13/17 km/h decreasing.

Thanks so much

I know they've already been answered, but..

1. where r is the diagonal distance;
r² = x² + 60²
2r.dr = 2x.dx
dr/dx = x/r
and dx/dt = 8
when r = 100, x = 80, so x/r = 4/5
dr/dx.dx/dt = dr/dt = 4/5 *8 = 6.4

2. dX/dt = 3, dY/dt = 4
R² = X² + Y²
2R.dR = 2X.dX + 2Y.dY
after 2 hours, the Y distance is 2*4 = 8, and the X distance is -21 + 6 = -15, so R = 17
17dR/dt = -15dX/dt + 8dY/dt
dR/dt = (-15*3 + 8*4)/17
= -13/17

 

[lyounamu]

haha.

Your tutorial was not at all brief. You handled the question exactly as how Cornoreos wanted it to be handled (as I'm using his textbook). It's just that I labelled my diagrams wrong and subbed y in a bit too early I just did about 5 of his questions and got them all right, so I'm very happy

Thanks again, Namu

They are from Coroneous? I thought they were from Fitzy as I saw the similar questions from there.

By the way, great work in getting those questions. Most of my accelerated friends wouldn't have got those questions correct.

 

[Doctor Jolly]

They are from Coroneous? I thought they were from Fitzy as I saw the similar questions from there.

By the way, great work in getting those questions. Most of my accelerated friends wouldn't have got those questions correct.

Yup they were from an orange book, titled 'Simplified HSC 3unit Course'. Oh yes, I flipped through Fitzy last night and I saw a similar question.

Thanks I hope that I remember how to do them once I get back to school though.