VCE Methods Question Thread!

[M_BONG]

Ok how on earth do you do this:

The volume of a cone is kept constant. If radius of base is increasing at 1% per second, find the percentage rate at which the height is decreasing.


Thanks

[lzxnl]

Ok how on earth do you do this:

The volume of a cone is kept constant. If radius of base is increasing at 1% per second, find the percentage rate at which the height is decreasing.


Thanks

Volume kept constant? Interesting.
V = 1/3 pi r^2 h. V is a constant so we have a relation between r and h
dr/dt = r/100 (that's what 1% increase means)
r^2 = 3V/(pi h)
2r dr/dh = -3V/(pi h^2)  =>   dh/dr = -2pi*r*h^2/3V
dh/dt = dh/dr * dr/dt
dh/dt = -2*pi*r*h^2/3V * r/100
= -2*pi*r^2*h^2/300V
Now r^2 = 3V/(pi h)
Sub that in => -2*pi*(3V / (pi*h)) h^2/300V = -h/50, which means there is a 2% decrease in h

lol wow. I didn't even know what I was doing there. Just use all the information given and hope something works.

[Mieow]

3) A and B are two points on a curve defined by f(x) = (x + 3)^3 and their x-coordinates are 1 and
1 + h respectively.
(a) Find the gradient of the chord AB.

The answer is h^2 + 12h + 48 but my answer is completely different. Any help is much appreciated 

[Zealous]

3) A and B are two points on a curve defined by f(x) = (x + 3)^3 and their x-coordinates are 1 and
1 + h respectively.
(a) Find the gradient of the chord AB.

The answer is h^2 + 12h + 48 but my answer is completely different. Any help is much appreciated 

The gradient of the chord AB will simply be the gradient of the line that connects the points A (1, f(1)) and B: (1+h , f(1+h)).





Our two points are now:



Substitute this into the equation of a gradient:



[edit] Here's f(x) represented graphically, you can drag the h slider around and see the corresponding value for m:
https://www.desmos.com/calculator/yygjqeovog

[keltingmeith]

f(1 + h) - f(1) / 1+h - 1
(h + 4)^3 - 4^3 / h
h^3 + 3h^24 + 3h4^2 /h
h(h^2 + 12h + 48)/h
h^2 + 12h + 48

Okay, so the formula for the gradient of a straight line (our chord is a straight line) is . So, our two x-coordinates are 1 and 1 + h, which means our two y-coordinates are f(1) and f(1 + h). So, this gives us the following equation:



After that, it's just a matter of number crunching.



Food for thought - what would you get if you then took the limit as h approaches 0?

[spectroscopy]

how to simplify:
(4x^4 y^2 ) ^2 + 2(x^2  y)^4

thanks

[spectroscopy]

also without using logarithms could someone please show me the steps to solving the following for x:  16^x =1024
cheers !

[soNasty]

hey spectroscopy, id just make the bases the same (if you dont wanna use logs)

so.. is the same as ..

then since the bases are the same you'd just solve for x

[spectroscopy]

thanks !

[spectroscopy]

sorry for the spam but what about if i were to try and solve log₂1/16 how would i go about doing that

[Mieow]

Thanks EulerFan and Zealous

I need help for one more question: For the function f(x) = x^2 − 4x, find the point of intersection between y = f(x) and y = f-1(x)(inverse). The answer is (5,5)

And how are you guys figuring out how to use the latex function ._.

[soNasty]

hey Mieow, with that function of x, to find the inverse, i'd first complete the square so it'd be easier.

so





then i would proceed to find the inverse of that function
i let f(x)=y and interchange x<=>y




therefore
hence

okay, so you know the equations reflect and hence intersect at the line y=x
so to make it easier, just make the inverse function equal to x, and solve for that



so we know that x = 0 and x = 5

well firstly, to have an inverse for x must be greater than two (thats where TP is) so we'll use x=5

so therefore we have the x value for the, so lets sub it in to either





therefore, intersection points are at (5,5)

[soNasty]

Spectroscopy, log2(1/16) is the same as log2(2^-4)
which is equal to -4 once you cancel the log out.

[Zealous]

how to simplify:
(4x^4 y^2 ) ^2 + 2(x^2  y)^4

sorry for the spam but what about if i were to try and solve log₂1/16 how would i go about doing that

Just index and logarithm laws.

I need help for one more question: For the function f(x) = x^2 − 4x, find the point of intersection between y = f(x) and y = f-1(x)(inverse). The answer is (5,5)
And how are you guys figuring out how to use the latex function ._.

The inverse function is the reflection of f(x) across the line y=x. Hence we can simply solve f(x)=x in order to find the intersection of f(x) and its inverse.



These are our x values for the point of intersection of f(x) and f-1(x).

Sub it back into f(x):

(or you could have subbed it into y=x, which would be easier...)

Therefore our intersection points are (0,0) and (5,5).

Here's f(x), f-1(x) and y=x represented visually. Here you can see that the intersection points of a graph and its inverse are on the line y=x.
https://www.desmos.com/calculator/wkrivzrqxk

[Cort]

Yeah, I have trouble identifying related rates of change -- I just have no idea which is which. Anyone have any tips on how to approach these?