VCE Specialist 3/4 Question Thread!

[Mieow]

how would I show that is a solution to the differential equation

[nhmn0301]

how would I show that is a solution to the differential equation

Diff arcsin(x/4)
dy/dx = 1/ sqrt ( 1 - (x^2/16) )       x 1/4
         = 1/ sqrt (16-x^2)    (sorry I skip some working out steps, but you get the idea)
d^2y/dx^2 = -1/2 (16-x^2)^(-3/2)  x (-2x)
                   = x/(sqrt 16 - x^2) ) ^3
Hence, LHS  = x/(sqrt 16 - x^2) ) ^3  - (1/ sqrt (16-x^2) )^3
                    = RHS

[liamhernon]

HI guys, need help with the following differential equation:
a and b are constants where x=t(acos2t + bsin2t) is a solution to the differential equation : (d^2 x)/(dt^2) + 4x = 2cos2t.
Find a and b.
Thanks in advance.

[keltingmeith]

I'm just going to skip the tedious steps, hope you don't mind. First step is to find x''. Some differentiation tells us that . Next, sub this into the equation, which gives us Some quick algebra reduces this to .

This is where the magic comes in - since this should work for all values of t, just sub in values of t and see what we can find for a and b!

For t = 0, we get 2b = 1, so b = 1/2. For t = pi/4, we get -2a = 0, so a = 0.

Subbing these into the original equation, we get LHS = RHS, so these are values for a solution to this differential equation.

[Bestie]

hello
can someone please help me with question 15?
please see attachment
thank you

[Ancora_Imparo]

15.a.
Find the equation of the tangent to the curve at point .

When :
Equation of tangent:




Sketch the graphs of and . The graph of is above the graph of . Thus, the area of the region bound by the curves and the y-axis is given by:
Area =

15.b.
Volume of solid of revolution of a region between two curves around the x-axis is given by:
Volume = , where f(x) is above g(x).

Thus:
Volume =

[Bestie]

what happened to the pi intergral x^2 dy?
i thought that was the formula to finding the volume of revolution around y axis?

[Ancora_Imparo]

Yes, that's correct, but I believe 15.b. is asking for the volume of revolution around the x-axis?

[Bestie]

oh whoops.. sorry i read the question wrong i thought it was asking for the volume around the y axis... sorry thank you for your help though

[Bestie]

can someone please help me on question 24b)
thank you

what ive done that i know should be correct is:
y^2 = 1/(x^2+9)
x^2 = 1/y^2 - 9

[Jacyan]

can someone please help me on question 24b)
thank you

what ive done that i know should be correct is:
y^2 = 1/(x^2+9)
x^2 = 1/y^2 - 9

See the attachment

[keltingmeith]

Step 1. Graph function and make sure to figure out what area you need. See attached for graph.
Step 2. Figure out what axis we're rotating around - in case 1 it's the x-axis, so we have a dx.
Step 3. Using your graph, figure out what your limits are. For case 1, it's made obvious we want between x=0 and x=4.
Step 4. Confirm you have everything you need, then apply the formula.



(cutting part b because I see I was beaten to it, darnit. )

[Bestie]

thank you both

but for the question 24b) how did you get 4^2 time 1/5 shouldn't the area of the rectangle be 8 time 1/5, how would i change it so that it rotates around?

[Zealous]

thank you both

but for the question 24b) how did you get 4^2 time 1/5 shouldn't the area of the rectangle be 8 time 1/5, how would i change it so that it rotates around?

Jacyan calculated the volume of a cylinder using where the radius was and the height was . Remember that we're dealing with volume here, not area - when we revolve a rectangle around an axis, we get a cylinder.

[Bestie]

oh ok yup thank you
i was just wondering why did i have to split it?