Related Rates rectangle area

[bondgirl007]

Homework Statement

A rectangle is expanding so that its length is always twice its width. The perimeter of the rectangle is increasing at a rate of 6cm/min. Find the rate of increase of the area of the renctangle when the perimeter is 40 cm.

Homework Equations

The Attempt at a Solution

p = 6x
dp/dt = 6 dx/dt
6 = 6 dx/dt
dx/dt = 1

40 = 6x
x = 40/6

a = 2x^2
da/dt = 4x^3 dx/dt
da/dt = 4(40/3)^3 (1)
da/dt = 32000/27

However, this is not the right answer as the correct answer is 80/3 cm^2/min.

Can anyone please help me solve this question? I have been stuck on it for the past half hour.

 

[rocomath]

$$A=2x^{2}$$

$$\frac{dA}{dt}=4x\frac{dx}{dt}$$

 

[bondgirl007]

Thank you so much! It worked!
Sometimes, small mistakes can make big differences.

 

[coomast]

You are on the right track. Keep in mind what you have and use that for the solution,
i.e. the derivative of the perimeter is known.
So in order to know the change in area with respect to time find the relation between the area and the perimeter.
Taking the derivative of this relation with respect to time gives you the equation you are looking for.

 

[coomast]

Strange, dx/dt is not given. (although dA/dt=4x*dx/dt is correct)
A=p^2/18, from which:
dA/dt=p/9*dp/dt
This is thus dA/dt=40cm/9*6cm/min=80/3cm^2/min

 

[rocomath]

Strange, dx/dt is not given. (although dA/dt=4x*dx/dt is correct)
A=p^2/18, from which:
dA/dt=p/9*dp/dt
This is thus dA/dt=40cm/9*6cm/min=80/3cm^2/min

you can solve for it, but it's been a while since I've seen Related Rates.

$$P=6w$$

$$\frac{dP}{dt}=6\frac{dW}{dt}$$

dP\dT given

$$6\frac{cm}{min}=6\frac{dW}{dt}$$

then i differentiated Area and plugged in and got 80/3.

 

[coomast]

OK, it's clear how you did it. It is a way bit longer to derive, but perfectly valid. Nice.

 

[bondgirl007]

Thanks everyone.

I have another question on related rates.

At 9am, ship A is 50km east of ship B. Ship A is sailing north at 40km/h and ship B is sailing south at 30km/h. How fast is the distance between them changing at noon?

I have drawn a diagram with two triangles connected but don't know how to get the distance between the two by differentiating a^2 + b^2 = c^2 with respect to t. Answer for this is 68 km/h. Any help would be GREATLY appreciated!

 

[ace123]

Which calc class are you in, meaning which college? If you are in college

Edit: Have any work for this btw?

 

[bondgirl007]

I'm not in college yet; grade 12 in high school.

Yes, I have homework on related rates and am struggling big time!

Any help on the above question?

 

[ace123]

I'm sorry it's been a while since I've done related rates so its taking me some time to get the answer you have

 

[arildno]

Thanks everyone.

I have another question on related rates.

At 9am, ship A is 50km east of ship B. Ship A is sailing north at 40km/h and ship B is sailing south at 30km/h. How fast is the distance between them changing at noon?

I have drawn a diagram with two triangles connected but don't know how to get the distance between the two by differentiating a^2 + b^2 = c^2 with respect to t. Answer for this is 68 km/h. Any help would be GREATLY appreciated!

Let B's inital position be (0,0), and A's position (0,50), measured in km.
Now, as a function of time, A's position A(t)=(40t,50), B's position B(t)=(-30t,0)

What is now the expression for D(t), the distance between the boats as a function of time?

 

[rocomath]

starting pos. is at 9am, you're trying to determine the change at noon. how many hours have passed and how far have you gone with respects to each ship?

 

[coomast]

The way to do this is to look for the equation stating the distance between the ships as a function of time. This can be done by considering the triangles you described.
The lengths are depending on time which is nothing more than the velocity times time.
L(t)=sqrt(4900t^2+2500)
This function needs to be differentiated with respect to time. The derivative is standard, a square root and a power. After this put in the numbers and there should your solution be.

In any case you should have 1470/sqrt(466)km/h, which is the correct one. It is approximately equal to 68km/h.

 

[bondgirl007]

The way to do this is to look for the equation stating the distance between the ships as a function of time. This can be done by considering the triangles you described.
The lengths are depending on time which is nothing more than the velocity times time.
L(t)=sqrt(4900t^2+2500)
This function needs to be differentiated with respect to time. The derivative is standard, a square root and a power. After this put in the numbers and there should your solution be.

In any case you should have 1470/sqrt(466)km/h, which is the correct one. It is approximately equal to 68km/h.

The answer is right but I'm still not confused as to how you got it.
Where did you get the equation for L(t)?

 

[arildno]

Have a look at my previous post.

 

[ace123]

It may be helpful for you to draw one big triangle to see the distance between the 2 boats per hour.

Edit: The boats are moving at opposite directions at different rates so you can som those to get the distance per hour, and you are given that the ships started out 50km apart. Get it now

 

[coomast]

The equation for the distance between the ships can be derived by considering the triangle formed by them. L=sqrt(a^2+b^2). What are the numbers a and b?

 

[bondgirl007]

I have been doing this question by deriving a^2 + b^2 separately for each ship.

Once for the south ship so for a I have 90km and da/dt is 30 km, b = 50 and db/dt = 0.

The second time I derive it, I derive it for the north ship so a = 120km and da/dt is 40km and b=50 and db/dt=0.

Where am I going wrong?

 

[rocomath]

you need the "hypotenuse"

 

[bondgirl007]

you need the "hypotenuse"

I have the hypotenuse as well. For the first one, I got root(10600) and for the second one, I got 130.

After getting the hypotenuse, I substitute everything in the differentiated a^2 + b^2 and add the two dc/dt for the two separate equations but that doesn't work. :(

 

[arildno]

The answer is right but I'm still not confused as to how you got it.
Where did you get the equation for L(t)?

Let B's inital position be (0,0), and A's position (0,50), measured in km.
Now, as a function of time, A's position A(t)=(40t,50), B's position B(t)=(-30t,0)

What is now the expression for D(t), the distance between the boats as a function of time?

I have been doing this question by deriving a^2 + b^2 separately for each ship.

Once for the south ship so for a I have 90km and da/dt is 30 km, b = 50 and db/dt = 0.

The second time I derive it, I derive it for the north ship so a = 120km and da/dt is 40km and b=50 and db/dt=0.

Where am I going wrong?

You can't do it separately!

Now, try first to establish the distance function!
You have been given the coordinates of A and B as functions of time; what must then the distance between them be, as a function of time?

 

[ace123]

Try making it as a one triangle so you can see that you can't do them seperately

 

[bondgirl007]

How do I get the distance function? I've only been taught to use Pythagoras for these questions.

Also, when I do it in one triangle, one side will be 50 but will the other be 210 as a result of adding 90km and 120km?

 

[coomast]

Hmm, the numbers a and b are the sides of the same triangle. Make a sketch at some time t. Ship B is at some position north and ship A somewhere south. Draw a horizontal line through ship B and a vertical one through A. The distance between the ships is L. Can you now derive the equation for L?

 

[bondgirl007]

Hmm, the numbers a and b are the sides of the same triangle. Make a sketch at some time t. Ship B is at some position north and ship A somewhere south. Draw a horizontal line through ship B and a vertical one through A. The distance between the ships is L. Can you now derive the equation for L?

I'm sorry but I'm still not getting it.

 

[ace123]

But they are going in opposite directions. So if it's 30 Km/hr in one direction and 40km/hr in the other. The total is going to be 70 km/hr correct?

 

[bondgirl007]

But they are going in opposite directions. So if it's 30 Km/hr in one direction and 40km/hr in the other. The total is going to be 70 km/hr correct?

I thought the total is 10km because they're in opposite directions.

 

[coomast]

The horizontal distance between the ships is constant and equal to 50km. Can you derive the vertical one?

 

[arildno]

How do I get the distance function? I've only been taught to use Pythagoras for these questions.

Also, when I do it in one triangle, one side will be 50 but will the other be 210 as a result of adding 90km and 120km?

Suppose you have two points in the plane, with coordinates (x,y) and (X,Y), respectively.
How do you find the distance between these two points?

 

[ace123]

it would be 10 if they were moving towards each other. If i am moving 30 ft/sec in one direction and you move 40 ft/sec in the other direction. The distance between us in 1 second would be what?

 

[bondgirl007]

When I put it in 1 triangle, here's what I get.

a = 210 km
da/dt = 10km

b = 50 km
db/dt = 0

c = square root (46600)
dc/dt = 9.73 after solving it but that is way off!

What am I doing wrong? :crying:

 

[bondgirl007]

it would be 10 if they were moving towards each other. If i am moving 30 ft/sec in one direction and you move 40 ft/sec in the other direction. The distance between us in 1 second would be what?

It worked after I put 70 in for da/dt!

Thank you so much, everyone!

 

[ace123]

I think only one of us should continue because all of us are saying the samething but in a different way and we end up confusing her. It shouldnt' be me because I haven't touched this topic in years

 

[rocomath]

It worked after I put 70 in for da/dt!

Thank you so much, everyone!

you're given dA\dt and dB\dt

let x be the distance in the x-axis

let x+y be the distance in the y-axis

$$x^{2}+(x+y)^{2}=z^{2}$$

you got your answer by luck, you're given dA\dt was not 70. i encourage you to keep working this problem!