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August 23, 2025, 09:01:47 am

Author Topic: VCE Methods Question Thread!  (Read 5636956 times)  Share 

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MoonChild1234

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Re: VCE Methods Question Thread!
« Reply #18450 on: March 15, 2020, 10:42:24 pm »
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hi, i tried to solve an inequality with circular functions on my ti inspire but it kept giving me general solutions even though i specified the domain, is there something i am doing wrong?

BiggestVCESweat

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Re: VCE Methods Question Thread!
« Reply #18451 on: March 16, 2020, 07:20:52 am »
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hi, i tried to solve an inequality with circular functions on my ti inspire but it kept giving me general solutions even though i specified the domain, is there something i am doing wrong?
Are you in radians mode?

MoonChild1234

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Re: VCE Methods Question Thread!
« Reply #18452 on: March 16, 2020, 05:44:19 pm »
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yeah i am

S_R_K

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Re: VCE Methods Question Thread!
« Reply #18453 on: March 16, 2020, 07:09:57 pm »
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Post what you have typed into your CAS and the inequality you are trying to solve.

MoonChild1234

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Re: VCE Methods Question Thread!
« Reply #18454 on: March 17, 2020, 05:27:11 pm »
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solve(sin(x)<0,x)|0<x<2pi

dream chaser

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Re: VCE Methods Question Thread!
« Reply #18455 on: March 17, 2020, 06:27:07 pm »
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solve(sin(x)<0,x)|0<x<2pi

Hi Moonchild1234,

To answer your question, you need to refer to the unit circle.

As sin(x) can be considered the y coordinate on the unit circle, sin(x)<0 when x∈ (π, 2π)

Hope this helps

Thanks,
dream chaser  :)

S_R_K

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Re: VCE Methods Question Thread!
« Reply #18456 on: March 17, 2020, 06:45:26 pm »
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solve(sin(x)<0,x)|0<x<2pi

So I also get max(0, 2pi*n – pi) < x < min(2pi, 2pi*n), where n is an integer. This only gives a non-empty interval if n = 1.

The Ti often has trouble with trig inequalities, not entirely sure why.

Geoo

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Re: VCE Methods Question Thread!
« Reply #18457 on: March 18, 2020, 11:53:32 pm »
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Dumb question....

Does f'(x) means that is the derivative (aka gradient function) and is f-1(x) means that it is the inverse function?
2020: VCE 93.2
2022: BSci/Arts (Chemistry/Pharmacology and French)@Monash

Sine

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Re: VCE Methods Question Thread!
« Reply #18458 on: March 18, 2020, 11:56:54 pm »
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Dumb question....

Does f'(x) means that is the derivative (aka gradient function) and is f-1(x) means that it is the inverse function?
Yes, that is correct :)

Azila2004

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Re: VCE Methods Question Thread!
« Reply #18459 on: March 20, 2020, 07:35:39 pm »
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Hello guys!

I've been trying to figure out how to get the correct answer in these problems. The Cambridge books only state the answers but not how to get them. There are the following:

1. The line y=mx+c is tangent to y^2 + (x-10)^2 = 25. Find m.
(This is the simplified form. I keep getting 1/3, but apparently it's incorrect. Help would be really appreciated!)

2. Let P(x1,y1) be a point on the circle x^2 + y^2 = a^2. Find the gradient of the line which is tangent to the circle at P.

3. An equilateral triangle ABC circumscribes x^2 + y^2 = a^2. The side BC has the equation x=-a. Find the equations of AB and AC, and hence the equation of the circle.
(Is there any topic I need to revise? I am not too familiar with this, or I may have forgotten.)


Thank you!

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S_R_K

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Re: VCE Methods Question Thread!
« Reply #18460 on: March 21, 2020, 01:42:25 pm »
+4
Hello guys!

I've been trying to figure out how to get the correct answer in these problems. The Cambridge books only state the answers but not how to get them. There are the following:

1. The line y=mx+c is tangent to y^2 + (x-10)^2 = 25. Find m.
(This is the simplified form. I keep getting 1/3, but apparently it's incorrect. Help would be really appreciated!)

There is not enough information to give a specific numerical answer to this question - are you given a point through which the tangent passes?

Quote
2. Let P(x1,y1) be a point on the circle x^2 + y^2 = a^2. Find the gradient of the line which is tangent to the circle at P.

A tangent to a circle at a point is always perpendicular to the diameter at that point. The diameter of the circle x^2 + y^2 = a^2, passing through P(x1, y1) on the circle, has equation y – y1 = (y1/x1)(x – x1). Hence the tangent passing through P(x1, y1) is y – y1 = –(x1/y1)(x – x1).

You can also set up a quadratic equation for the point of intersection, and then set the discriminant equal to zero to find the gradient.

Quote
3. An equilateral triangle ABC circumscribes x^2 + y^2 = a^2. The side BC has the equation x=-a. Find the equations of AB and AC, and hence the equation of the circle.

Each angle in an equilateral triangle is 60°. Since the side BC is perpendicular to the x-axis, the other sides must form angles of 150° from the positive x-axis and -150° from the positive x-axis, respectively. Then use m = tan(θ) to find the gradients of the sides AB and AC. You can also use trigonometry to find the coordinates of A, B, C, and then the equations of the lines follow.

Quote
(Is there any topic I need to revise? I am not too familiar with this, or I may have forgotten.)


Knowing some circle geometry is often useful when dealing with questions about intersecting lines and circles.

TheEagle

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Re: VCE Methods Question Thread!
« Reply #18461 on: March 21, 2020, 06:56:34 pm »
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Can someone please help me
thanks

Sine

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Re: VCE Methods Question Thread!
« Reply #18462 on: March 21, 2020, 07:03:59 pm »
+1
Can someone please help me
thanks
What methods have you thought about?

Can you connect it to any more simpler concepts?

TheEagle

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Re: VCE Methods Question Thread!
« Reply #18463 on: March 21, 2020, 07:15:50 pm »
+1
What methods have you thought about?

Can you connect it to any more simpler concepts?

Thus far, I have considered the following:

m=tan (theta)
therefore
theta=arctan(m)

I have also recognised the acute angle between the lines. My main issue is getting to the angle. I also realise that theta is the angle made with the positive x axis


TheEagle

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Re: VCE Methods Question Thread!
« Reply #18464 on: March 21, 2020, 08:32:57 pm »
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Got it!

My calculator was in radians haha :)