jack_chay's methods question thread

[jack_chay]

thank you darklight

[jack_chay]

The volume, V cm3, of right circular cone of height h cm and
radius r cm is given by V = 13
r2h and the curved surface
area, S cm2, of the cone is given by S = r * (r^2 + h^2)^1/2
Sand falls onto a horizontal floor at a rate of 20 cm3/s.
The sand falls in a pile so that it is in the shape of a right circular
cone with vertical angle 60◦. The height of the sand t seconds
after the sand starts to fall is x cm.

how to find dx/dt in terms of x?

I know r = (1/((3)^1/2) ) * x

[silverpixeli]

The volume, V cm3, of right circular cone of height h cm and
radius r cm is given by V = 13
r2h and the curved surface
area, S cm2, of the cone is given by S = r * (r^2 + h^2)^{1/2}
Sand falls onto a horizontal floor at a rate of 20 cm3/s.
The sand falls in a pile so that it is in the shape of a right circular
cone with vertical angle 60◦. The height of the sand t seconds
after the sand starts to fall is x cm.

how to find dx/dt in terms of x?

I know r = (1/((3)^1/2) ) * x

Very similar to the last one, this is another related rates problem (rates problem where you have one rate and need another)

What we need is dx/dt

What we have is

So we need to construct a chain rule equation that has dx/dt in terms of dV/dt and something else, then we find the something else and we'll be able to find dx/dt

(chain rule)

so the other rate we need to find first is dx/dV, meaning we need a rule linking x and V (and we can differentiate this)

lucky we have and h=x and r= (1/((3)^{1/2}) ) * x
so we can sub those values in to get
which simplifies to  (you should check my working to make sure this is right, I did it quickly)

from that we know that (from differentiating)
but we need dx/dV, not dV/dx so we need to flip our answer to (units of length/units of volume)

NOW we are getting somewhere, we went through all that work to get dx/dV so we can go back and sub it into the rule from the start


simplify along with units to give
, finally arriving at dx/dt in terms of x

if the bit with the units confused you, they don't really matter, it's just a good way to check you have found the right rates, I'm not really sure how to explain the dimensional analysis behind it, but for dx/dt which is length/time you should end up with something like cm/sec which is what we did get

the key thing to take away from this question is that when we need one rate and we have another, we can use the chain rule to figure out the required rate with the help of another two rates (one of the two we are usually given, the other one we have to find, usually by differentiating something)

hope that helps

EDIT: I'm pretty sure the volume of a cone has a pi in the formula.. I just used your numbers, but you'll want to check that up. Volume of a cone; V=1/3 * pi * r^2 * h

[b^3]

EDIT: I'm pretty sure the volume of a cone has a pi in the formula.. I just used your numbers, but you'll want to check that up. Volume of a cone; V=1/3 * pi * r^2 * h

Yeah that's it. It's on the formula sheet, but if you got bored in spesh and wanted to derive it, you could use the volume of revolution method (and yeah I know this is a methods thread, hence putting it in the spoiler below, so don't worry about this methods-only people).

Spoiler

Firstly lets look at the slanted edge of the cone. What equation represents this line? The points and lie on the line, and we know that it's a straight line. So
for
Now if we rotate this line around the axis, we will form a cone. We can find the volume of this cone by working out the volume of revolution formed. So for a volume of revolution around the axis we have:

So we need in terms of , rearranging gives:

Now we are rotated from to , so our lower and upper terminals are and respectively. Substituting in and integrating gives

That is we get back our formula for the volume of a cone. Anyways, that's my mathematical tangent for the day

[jack_chay]

thank you so much everyone

[jack_chay]

g : (1,∞)→R, g(x) = 1/(x − 1)
The line segment AB is drawn from
the point A(2, 1) to the point B(b, g(b))
where b > 2.

At what value of x between 1 and b does
the tangent to the graph of g have the
same gradient as AB?

I know the gradient of AB is 1/(1-b). But that's all.

help...

Thank you in advance

[Phy124]

Gradient of AB is

We want to find when the gradient of g(x) is equal to this value.









Although

So 

[jack_chay]

thank you

[jack_chay]

i'm not sure how to set out this show it question.

The diagram below shows the graph of the function f: R+ →R, f (x) = 1/ x^2 .
The line segment AB is drawn from the point A(1, 1) to the point B(b, f (b)) where b > 1.

Show that for b > 1,
                                     b
(insert integral line)     1 f (x) dx < 1.

[b^3]

[jack_chay]

im sorry im confused on how you got the second part

[b^3]

Here's the thought process for the working.

Firstly lets look at the fraction . We have the restriction . What happens if is really close to 1 but only slightly larger? Then the fraction will be really close to 1 itself. But we can make really big, so if we let , then the denominator of the fraction becomes really large, so the fraction itself goes to zero, . So we know that the fraction has to be between these two values, . (You'd normally also check that the curve doesn't do anything funny/weird between 1 and , but since we know the shape of the curve here we know it doesn't).

Now if we know that , then what values can take? It will be able to take the negative of the values it took before, so .

Now for the last bit. We are going to subtract a value between 0 and 1 form 1. (Now that I think about it, you could probably do it without the previous paragraph in there). We're taking a function that has range away from a constant. So the maximium we can get is (non-inclusive) and the minimum we can get is (non-inclusive). Since we are only asked to show that the integral is less than one, we will only need the first part, .

Hope that helps

[jack_chay]

yup i got it now thank you

Edit 1: now that i actually did it, i'm sorry but what was that last paragraph about? the one about subtracting a number between 0 and 1? I'm sorry...

Edit 2: on second thoughts, i got it thank you !!!

Edit 3: i'm a little stuck...

Please Help

if f((2x+ 3)/x)= 〖(x+3)〗^2
if f(x) domain is restricted to (- infinity, -1], what is the restricted domain of f((2x+ 3)/x)?

Thank you

Mod Edit: Merged triple post (Please use the edit button in the future, it's there for a reason )- Snow Red

Mod Edit 2: Merged another post. Jack pls. - Tim