VCE Methods Question Thread!

[Rishi97]

An inverted right circular cone is filled with liquid. The cone has a radius of 3m and height of 7m. The liquid flows from the apex of the cone at a constant rate of 0.6m³/min. Find the rate at which the depth of the liquid is dropping, when the depth of the liquid is 2m.

[Zealous]

An inverted right circular cone is filled with liquid. The cone has a radius of 3m and height of 7m. The liquid flows from the apex of the cone at a constant rate of 0.6m³/min. Find the rate at which the depth of the liquid is dropping, when the depth of the liquid is 2m.

"The liquid flows from the apex of the cone at a constant rate of 0.6m³/min."

(note the negative because it is flowing out!)

"Find the rate at which the depth of the liquid is dropping, when the depth of the liquid is 2m."



Set up your chain rule equation:



Now we need to find what dh/dV is. Let's first find dV/dh, then we can flip it to get dh/dV. We know that it is a right circular cone so it follows that:

- but we cannot differentiate this straight away in terms of 'h' because there is another variable 'r' in the equation, so we need to eliminate the 'r'. So this is where similar triangles comes in:



    We have found what 'r' is in terms of h, so let's sub this back into V:

    Now let's differentiate it with respect to 'h' and flip it:



Bring this back into the chain rule equation and substitute in our given condition where h=2:





The depth of the liquid is decreasing at a rate of when the depth of the liquid is 2m.

[Frozone]

Could someone please show me the steps for question c,b,d. My calculations are "just" missing the answer.

[Mieow]

Given that and when x=1, and y= 2, find y when x=0.

Could someone please help me with that question?

[Zealous]

[Thorium]

Could someone please help me with that question?

Sub x=1, dy/dx=3 and solve for a. Then, untidifferentiate the expression (and dont forget the "+c" at the end). Now you have found the equation needed, but it is not complete, i.e. You dont know value of c. So you can sub in x=1, y=2 to find c. Now that you have your y equation, you can answer the question asked ("find y when x=0"). So all you need to do is to let x=0 and find the value of y. That will be the answer.

Sorry i was too abstract, but hope it helps.

EDIT: zealous, i think you forgot the very last step
EDIT: yes, you noticed it.

[Mieow]

Thanks Zealous and Thorium!

[lzxnl]

Fun fact: if you have CAS, you can solve the differential equation in one go by setting up a definite integral.

 



Note how I changed the integration variable to t to let x be the upper bound? The x=1 condition appears in the bottom terminal of the integral and the y=2 condition appears outside the integral. Try it. This will give you the same answer.

Rationale? If dy/dx = f(x) and f(a) = b, then the integral between a and x of f(t) should give the change in y from when t=a and t=x (a variable)
But the change in y is y (at any given x) - b, which is y when t = a
Thus y - a = integral of f(t) dt from t=a to t=x

[Rishi97]

The curve with equation y= x³ + ax² +bx + c passes through (0,7) and has a stationary point at (2,15). Find a, b and c
Thanks

[keltingmeith]

We have three unknowns, so we need three equations. (0, 7) is a point, giving us 7 = c, so now we just need two equations.

There's a stationary point at (2, 15) - so not only dies f(2) = 15, but f'(2) = 0. So, 15 = (2)^3 + a(2)^2 + 2b +7 and 0 = 3(2)^2 + a(2) + b

Now, just solve for a and b.

[IndefatigableLover]

The curve with equation y= x³ + ax² +bx + c passes through (0,7) and has a stationary point at (2,15). Find a, b and c
Thanks

Not sure if I did this right but you sub in (0,7) into the main equation and you'll get c=7.

From there you derive the main equation and you should get
From here you know that you have a SP at (2,15) so we'll sub in x=2 into here and from here you can rearrange to make 'a' the subject.



Now back to the main equation you resub (2,15) and rearrange till you get and then you substitute 'a' into here where you'll find that 'b' = 12. Now that you know that b=12, you just sub that back in and you'll find that 'a' = -6.

So your final equation is:



https://www.desmos.com/calculator/de2vkcib0m

Appears you have a Stationary POI at (2,15)... is that right? ._.

[Rishi97]

We have three unknowns, so we need three equations. (0, 7) is a point, giving us 7 = c, so now we just need two equations.

There's a stationary point at (2, 15) - so not only dies f(2) = 15, but f'(2) = 0. So, 15 = (2)^3 + a(2)^2 + 2b +7 and 0 = 3(2)^2 + a(2) + b

Now, just solve for a and b.

Perfect got the answer thanks

[Rishi97]

Not sure if I did this right but you sub in (0,7) into the main equation and you'll get c=7.

From there you derive the main equation and you should get
From here you know that you have a SP at (2,15) so we'll sub in x=2 into here and from here you can rearrange to make 'a' the subject.



Now back to the main equation you resub (2,15) and rearrange till you get and then you substitute 'a' into here where you'll find that 'b' = 12. Now that you know that b=12, you just sub that back in and you'll find that 'a' = -6.

So your final equation is:

You derived the right answerss . lol
Thanks

[keltingmeith]

Definitely works, indefatigable. My advice - don't doubt yourself. More often than not, if you think you're right, you're right. Especially in a SAC or exam, you don't have time to second guess yourself anyway, so just go in strong, and believe you've done it right. You can doubt yourself after everyone's done something different to you.

[Rishi97]

A piece of wire 100 cm in length is bent into the shape shown
a ) Find y in terms of x (I got y=50-25x )
b ) Find the area of the shape A cm², in terms of x

Its a really simple questions but I keep stuffing up the working out for area. I don't know what I'm doing wrong so it's really frustrating
Pls help