VCE Methods Question Thread!

[Quantum44]

(Image removed from quote.) number 5

Sub in x=3, y=50 to get equation 1
Sub in x=6, y=10 to get equation 2

Solve both equations simultaneously for variables a and b

[captkirk]

Help

If f'(x)=acos (ux)-be^vx

and

f(x)=sin(3x)-3e ^-3x
-1 

find the exact values of a,b,u,v

[LPadlan]

Sub in x=3, y=50 to get equation 1
Sub in x=6, y=10 to get equation 2

Solve both equations simultaneously for variables a and b

Yes, i know that. But i dont know how to solve after i make the two equations. If you dont mind could you show me a step by step solution? Thanks

[Sirius]

Help

If f'(x)=acos (ux)-be^vx

and

f(x)=sin(3x)-3e ^-3x
-1 

find the exact values of a,b,u,v

Differentiate f(x)=sin(3x)-3e^-3x -1 
You get f'(x)= 3 cos(3x)) +9*e ^-3x   

Therefore, a= 3 b= -9  u= 3  and v= -3

[Sine]

Yes, i know that. But i dont know how to solve after i make the two equations. If you dont mind could you show me a step by step solution? Thanks

Hopefully this helps you! 

[LPadlan]

7 with step by step solutions pls

[Guideme]

Pls help me with question 12 pls thank you

[Syndicate]

May someone explain how to do this question?

Volume of a cylinder = \(\pi r^2 h\)

a) r = 6 V1 = \(36 \pi h\)
Domain of V must be all real positive numbers

(Image removed from quote.) 7 with step by step solutions pls

1. Write down your two simultaneous equations.
coordinates 1:

10 = alog_2( + b
10 = 3a + b (1)
14 = alog_2(32) + b
14 = 5a+b (2)

(2)-(1)
=> 4 = 2a
therefore a = 2
Sub a = 2 into (1)
10 = 6 + b
therefore b = 4

Pls help me with question 12 pls thank you (Image removed from quote.)

12a) distance (AP) = 300/cos(x)      (where x is theta)
time = distance/ speed => 300/4cos(x) = 75/cos(x)

b) distance(PC) = distance(BC - BP)
= 1100 - 300/tan(x)
time = (1100 - 300/tan(x))/ 5 = 220 - 60/tan(x)

c) Total time = time(AP) + time(PC)
= 220 - 60sin(x)/cos(x) + 75/cos(x)
= 220 + (75-60sin(x))/cos(x)

d) dT/dx = 75sin(x)/cos^2(x) -60sex^2(x)  (you can do this by hand using the quotient rule)

e) Set dT/dx = 0

therefore 75sin(x) - 60 = 0
sin(x) = 4/5
x = 53 degrees approx.

f) Min time occurs at x = 53 degrees
T = 75sin(53.13) -60/cos^2(53.13) = 265 seconds

To find distance of BP, BP = 300/tan(53) = 400 metres

[lillianmaher]

Does anybody have any questions/ examples/ worked answers for questions from Unit 3 involving a function, containing three unknown constants within the equation for which you have to solve for? You are required to make assumptions and justify these assumptions in order to obtain the values of the constants to be able to graph an 'appropriate' function. Please if anyone has any of these types of questions could you please teach me how to do them?
Thanks

[captkirk]

A) and c) by the way did I do b correct? Thanks
http://imgur.com/a/kp32f

[captkirk]

Another one
Determine when

∫(e^2xsin(2x))dx=∫(e^2xcos(2x))dx, given that 0≤x≤2π

[Azzzz]

Going by the question itself and how it's worded, the equation should be in the form of y=mx+3.9, where m is the gradient between 6 and 24 weeks.

Ahhh okay thank you so much!

[Azzzz]

The domain is only affected by the translation of x units along the x-axis, or being dilated across the y-axis. I hope someone can help with your your TI-nspire question (I am not the best at using it either).

Firstly, write down all the information you know.
t (days) = 0, w (kilograms) = 3.9
From this you should already be able to tell that c = 3.9
m = the rate of change in his weight and age
m = (8-5.5)/126
= 5/252

So therefore w = 5/252 t + 3.9
at t = 40 x 7
w = 5(40 x 7)/252 +3.9
w = 9.46 kg

I used days as if I set my y variable in weeks, the points would be inconsistent with the line. So therefore you are correct. The method you used is essentially what your friends did, however, they disregarded the initial weight information which would have led them to getting the wrong answer.

EDIT: I'll post my solution anyways as I was in the process of doing it when Yueni posted his 

Very clear explanation thank you very much Syndicate! As helpful as always

[The Waterboy]

Halo! For my attached question part d, iii) based of VCAA's definitions of increasing, we would use inclusive brackets around the extreme values of the domain ay? So [0, 5.21] http://imgur.com/a/3qAn1

[camrenis]

A) and c) by the way did I do b correct? Thanks
http://imgur.com/a/kp32f

Hints:
a) Find the length of the sides of the box with respect to \(x\), and the volume is the length squared multiplied by the height.
b) Almost, but physically can \(x\leq0\) and \(x\geq\frac{5}{2}\) if you want at least some volume?
c) Derive \(V=4x^3-20x^2+25x\) with respect to \(x\) and solve for setting the derivative to 0 (min/max).  Choose the value of \(x\) within your domain found in b), and hence substitute back into \(V\) to find the volume.

Asnwers:

Spoiler

a) As you are essentially cutting the corners out and bending the sides up, the length of each side becomes \(5-2x\), as two corners (i.e. 2 of \(x\)) has been cut out.  It's obviously a square, so the area of the square is then $$A=(5-2x)^2$$
The height of the box is simply \(h=x\) as the sides getting bent upwards to form the box have a width of \(x\).  The volume is therefore
$$
\begin{align}
V&=h\times A\\
&=x(5-2x)^2\\
&=x(25-20x+4x^2)\\
&=4x^3-20x^2+25x
\end{align}
$$

b) You pretty much had this, but values below 0 and above 2.5 can mathematically exist in the equation, but can't exist physically, so you are restricted to \(x\in\left[0,\frac{5}{2}\right]\).  Assuming we want at least some volume, then \(x\in\left(0,\frac{5}{2}\right)\).

c) Take the derivative of the volume function, so
$$
\begin{align}
\frac{dV}{dx}&=\frac{d}{dx}(4x^3-20x^2+25x)\\
&=12x^2-40x+25
\end{align}
$$
Set this to 0 to find the stationary points of the cubic function (i.e. either a minimum or maximum).  This can be solved many ways, you could try and factorise the above equation, but I'll just use the quadratic formula as it's what I prefer to use.
$$
\begin{align}
x&=\frac{40\pm\sqrt{40^2-4\times12\times25}}{2\times12}\\
&=\frac{40\pm\sqrt{400}}{24}\\
&=\frac{40\pm20}{24}\\
&=\left[\frac{5}{6},\frac{5}{2}\right]
\end{align}
$$
Only \(x=\frac{5}{6}\) is within our domain in b), so then substituting this into \(V\) gives
$$
\begin{align}
V&=4\left(\frac{5}{6}\right)^3-20\left(\frac{5}{6}\right)^2+25\left(\frac{5}{6}\right)\\
&=\frac{250}{27}\\
&\approx9.26
\end{align}
$$
which is the maximum volume.