HS Calculus question.. help pls?

[LuDaCriS66]

Just have a test coming up on velocity, acceleration, related rates, and newton's method... I kind of understand some of the chapter.. like for velocity and the such which I'm pretty sure is mostly taking the derivative of the function to find each thing..

for this kind of question... i'm lost.. it's almost like geometry and I've never been able to do geometry

1) A paper cup has the shape of a cone with height 8cm and a radius 3cm at the top. Water is poured into the cup at a rate of 2 cm^3/s.
How fast is the water level rising when the water level is 6cm deep?

If you can help, please explain how I should do this in detail.. I'm a complete idiot when it comes to math...

I don't even see how this question is similar...

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

3) 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 rectangle when the perimeter is 40cm.


Oh and I assume that this question is to be answered with newton's method right?

4) Find the root of the equation x^5= x + 2 correct to six decimal places.

I guess I just have to keep going until I get the same result to 6 decimal places two times in a row?

Anyway.. don't do all the questions.. unless you really want to of course. I just need a basic understanding of how to do those top 3 questions.

Thanks

 

[kt]

I am not sure if this is right or not, someone correct me if I am wrong, but for the first question:

You know the rate at which the water is filling up the cone (2 cm^3/s). What they are asking for is the rate cm/s. If you take the derivate of 2 cm^3/s going from 0 to 6, you'll get the rate cm^2/s. Now, all you have to do is find the radius of the cone at that height (6 cm) and you'll get your answer.

Second and third question are similar. Second question, you'll need to find how fast the vector between the two ships are changing. From the information given, you know the vertical distance between ship A and B is changing at a combined rate of 70km/h (40km/s + 30km/h). The horizontal distance remain constant at 50km and changing rate is (0 km/h). After 3 hours, vertical distance is 210km, horizontal distance is 50km, and using Pythagorean theorem diagnol distance between the ships is around 216km. Now, just do the math

 

[LuDaCriS66]

Thanks for the info. Anyone else have any input??

 

[Fandu]

I get dH/dt = .1257cm/s

First, you know the formula for the voluse of a cone, V=[pi]r^2h/3
Now insteat of having r in there, we can just relate that to the height. We know the cone is 8cm tall and 3 wide, so r=3h/8

Sub that into the volume formula, get V = (3h/8)^2 * [pi]h/3

Now take the derivative of that formula, now we already know h is 6cm, and dV/dt is 2cm^3/s. Rearrange that and get dH/dt = 32/(81[pi])

 

[Fandu]

For the second one:

The ships are 50km apart, and moving perpendictularly apart from each other at 70km/h.

So draw you two ships horizontally apart from each other, 50km. Now draw your 40km/h north vector on ship A, and a 30km/h south vector on ship B.

Now instead of using the original horizontal line as a base, draw a horizontal line at the tip of the north vector, back to perpendicular with ship B. So now we can see a right triangle with one side 50km and the other given by 70tkm. I'll use 't' for time in hours.
So the hypotenuse is:

D^2 = 50^2 + 70t^2

So now differentiate that with respect to t, and get:

dD/dt = 9800t / 2sqrt(50^2 + (70t)^2)
(I skipped a few algebra steps)

So we started this at 9am, and we want at noon, so that's 3 hours, so t=3, dD/dt = 68.09km/h


EDIT: My mistake was that I took 70t^2, instead of (70t)^2. This answer looks correct.

 

[Fandu]

Third one:

You have a rectangle with sides x, and 2x. So the perimeter of it is 6x. (P=6x) and the area of it is 2x^2 (A = 2X^2). Now we need to relate the perimeter and area, so put perimeter in terms of x (x = P/6), and stick that into our area formula to get:

A = 2(P/6)^2

We already know that the perimeter is increasing at a rate of 6cm/sec, so define dP/dt = 6
Differentiate A, with respect to t in minutes. and rearrange to get:

dA/dt = (4P*(dP/dt)) / 36

We know the perimeter is 40cm, so put in P=40, and get an answer of 26 2/3 cm^2/min

 

[kt]

<< For the second one:

The ships are 50km apart, and moving perpendictularly apart from each other at 70km/h.

So draw you two ships horizontally apart from each other, 50km. Now draw your 40km/h north vector on ship A, and a 30km/h south vector on ship B.

Now instead of using the original horizontal line as a base, draw a horizontal line at the tip of the north vector, back to perpendicular with ship B. So now we can see a right triangle with one side 50km and the other given by 70tkm. I'll use 't' for time in hours.
So the hypotenuse is:

D^2 = 50^2 + 70t^2

So now differentiate that with respect to t, and get:

dD/dt = 70t / sqrt(50^2 + 70t^2)
(I skipped a few algebra steps)

So we started this at 9am, and we want at noon, so that's 3 hours, so t=3, dD/dt = 3.75km/h
That looks kinda low...... but I don't see any mistakes that jump out. >>


Hmm.. I could be wrong, but 3.75km/h doesn't sound right. If you think about it, their rate of separation in vertical line is 70km/h, their vector distance should be moving at a faster rate than 70km/h.

 

[Fandu]

Fixed, see above.

Actually, it can never be more than 70km/h. Think about this scenario: If you have a triangle with sides 4m, and 4m. Now if you increase 1 side to 5, the hypotenuse only increases by ~.7.

In your problem, the north-south distance is increasing at 70km/h, but we want the diagonal distance. If you want to see the vectors, lets setup the scenario at noon:

the ships are 210km apart north-south, and still 50km apart east-west. So draw a 210km vectorN-S, and from the tip of that, one 50km east. What your saying (in essence) is that the diagonal is more than the magnitude of the 2 vectors added, but it's always less.

 

[LuDaCriS66]

Wow, thanks!! I really appreciate the work you put into that.. thanks so much!

 

[Martin]

What Fandu said


For rates, the general way goes like this: find function (equation) linking variables, express function with 1 variable, differentiate, plug in numbers.

For optimization: Find function, get to 1 variable, differenitate, set equal to zero, solve for X, plug in X and endpoints in to function.




You just need practice.

 

[kt]

<< Fixed, see above.

Actually, it can never be more than 70km/h. Think about this scenario: If you have a triangle with sides 4m, and 4m. Now if you increase 1 side to 5, the hypotenuse only increases by ~.7.

In your problem, the north-south distance is increasing at 70km/h, but we want the diagonal distance. If you want to see the vectors, lets setup the scenario at noon:

the ships are 210km apart north-south, and still 50km apart east-west. So draw a 210km vectorN-S, and from the tip of that, one 50km east. What your saying (in essence) is that the diagonal is more than the magnitude of the 2 vectors added, but it's always less. >>


Yeah, you're right.. haven't touched math for at least 8 years. I am thinking of the magnitude and not the rate of change

 

[alee25]

for the first question i got something different from fandu i got 1.1317 cm/sec, which sounds a bit more reasonbale than his answer

 

[LuDaCriS66]

well I checked the answer for that question in the back and it says this..

32/81(pie) =(with a dot on the top) 0.13 cm/s

so i guess Fandu was right

 

[alee25]

<< I get dH/dt = .1257cm/s

First, you know the formula for the voluse of a cone, V=[pi]r^2h/3
Now insteat of having r in there, we can just relate that to the height. We know the cone is 8cm tall and 3 wide, so r=3h/8

Sub that into the volume formula, get V = (3h/8)^2 * [pi]h/3

Now take the derivative of that formula, now we already know h is 6cm, and dV/dt is 2cm^3/s. Rearrange that and get dH/dt = 32/(81[pi]) >>




i agree up to the V=(3h/8)^2 * pi*h/3 which aslo is equal to V= 3(h^3) [pi]/64
so you take the dV/dT of that function and you get 3pi/64 *2h*dh/dt (remeber chain rule & implicit derivitive on the h^3)

so then you get the equation, 2cm/sec = 3pi/64*(2x6)*(dh/Dt) solve for dh/dt
which gives you 1.1317 cm^3/sec





EDIT nevermind i guess i was wrong. what did i do wrong then?

Edit: ahh figured it out, i do get .125 now, it should be:

i agree up to the V=(3h/8)^2 * pi*h/3 which aslo is equal to V= 3(h^3) [pi]/64
so you take the dV/dT of that function and you get 3pi/64 *3h^2*dh/dt (remeber chain rule & implicit derivitive on the h^3)

so then you get the equation, 2cm/sec = 3pi/64*(3x36)*(dh/Dt) solve for dh/dt

 

[Fandu]

V = 3[pi]h^3 / 64

Take the derivative, you get:

dV/dt = ((9[pi]h^2) / 64) * dH/dt

Now sub in dV/dt = 2, and h = 6, so then you have:

2 = 9[pi](6^2) / 64 * dH/dt

Take the LHS, multiply by 64, divide by 9[pi](36), so re-write that again:

128/324[pi] = dH/dt

which reduces to 32/81[pi]

In my books, that's an exact answer

 

[alee25]

fandu, wanna explain more on how you did number 2? i can do number 3, but mabye its the wording of number 2 but i cant seem to get it

 

[Fandu]

OK, initial scenario,

So, now we can re-arrange it so that we only have velocity on one side of our triangle that we setup:

re-arrange

So, now 3 hours later we have this setup.

Have you followed up to here? So now did you follow what I did above?

 

[alee25]

ahhh yes i see! thx
i just didnt get how you moved both velocites to one speed and placed that on the triangle