A few quick questions: graphing, circle geometry, trigonometry (1 Viewer)

[bawd]

Appreciate the help, and please show the working out and a brief explanation, as I'm a bit stumped for these questions.

1. Find the important features of y = (x - 1)/(x^2 - 9)

First, I know the x asymptotes for this question is x = -3, 3. I've also differentiated the equation as such:

For stationary points,

y' = (x^2 - 9 - 2x(x - 1)) / (x^2 - 9)^2 = 0

= (x^2 - 9 - 2x^2 + 2x) / (x^2 - 9)^2 = 0

= - x^2 - 9 - 2x = 0

But then I get stuck and can't solve it for x. Also, how do you find the horizontal asymptotes and other important points for this question?

2. The sum of the radii of two circles is 100 cm. If one of the circles has a radius of x cm, show tat the sum of the areas of the two cirlces is given by: A = 2pi (x^2 - 100x +5000). Find the value of x for which A is the least.

3. Prove the circles x^2 + y^2 + 4x - 10y + 20 = 0, x^2 + y^2 - 12x + 2y = 12 touch externally.

4. Solve sin2A = sin A, 0 <= A <= 360. Also find the general solution.

Thanks.

 

[shaon0]

Appreciate the help, and please show the working out and a brief explanation, as I'm a bit stumped for these questions.

1. Find the important features of y = (x - 1)/(x^2 - 9)

First, I know the x asymptotes for this question is x = -3, 3. I've also differentiated the equation as such:

For stationary points,

y' = (x^2 - 9 - 2x(x - 1)) / (x^2 - 9)^2 = 0

= (x^2 - 9 - 2x^2 + 2x) / (x^2 - 9)^2 = 0

= - x^2 - 9 - 2x = 0

But then I get stuck and can't solve it for x. Also, how do you find the horizontal asymptotes and other important points for this question?

2. The sum of the radii of two circles is 100 cm. If one of the circles has a radius of x cm, show tat the sum of the areas of the two cirlces is given by: A = 2pi (x^2 - 100x +5000). Find the value of x for which A is the least.

3. Prove the circles x^2 + y^2 + 4x - 10y = 0, x^2 + y^2 - 12x + 2y = 12 touch externally.

4. Solve sin2A = sin A, 0 <= A <= 360. Also find the general solution.

Thanks.

1) f'(x)=x^2-2x-9
Let f'(x)=0
x^2-2x-9=0
delta=b^2-4ac
=4-4*1.-9
=40
x_1,2 = 2+-sqrt(40)/2
= 1+-sqrt(10)
x_1 = 1+sqrt(10) and x_2 = 1-sqrt(10)
ie. Stationary points

4)
sin2A=sinA
2sinAcosA=sinA
2cosA=1
cosA=1/2
A= 60, 300

I'll do the rest if i have time later.

 

[gurmies]

For question 1:

Do what shaon0 said. In addition, find x and y intercepts. y = (x - 1)/(x^2 - 9), Thus when x = 0, y = 1/9 Now when y = 0 , x = 1. The y-intercept is (0, 1/9) and the x - intercept is (1, 0). Now you say you're having problems finding horizontal assymptotes. There is quite a simple way of doing this, which has never failed me. If the denominator has a higher power of x than the numerator, the horizontal assymptote will be y = 0 (in other words, the x - axis). If the denominator has an equal highest power to the numerator, simply cancel these leading coefficients to get your assymptote. For instance: (x^3 + x^2)/(3x^3 + 1), the highest power of numerator and denominator is to the 3rd degree. We cancel the coefficients, and y ----> 1/3, as x ------> infinity. Now the last case is the trickiest type, usually encountered in the extension II course, where the numerator contains a higher power of x than the denominator. An oblique assymptote comes from this situation. In the first two examples we have been given a horizontal assymptote, i.e y = m, m integral. In the last case, you divide top by bottom to get your oblique assymptote. This isn't very important for extension 1, so don't worry too much about it.


For question 2:

If radii add to 100, and one radius is x, therefore the other must be 100-x. Taking both areas:

A1 = pi (x^2)

A2 = pi ((100-x)^2)

= pi (10000 - 200x + x^2)

Now A1 + A2 = pi ( x^2 + 10000 - 200x + x^2) -----> Factoring pi out

= pi (2x^2 - 200x + 10000)

= 2pi (x^2 - 100x + 5000) ----> Factoring 2 out

Now, 2pix^2 - 200pix + 10000pi. Since we want a minimum value, we need to find the vertex.

x value of the vertex is given by -b/2a.

(200pi)/4pi = 50.

Therefore, when x = 50, the minimum area is achieved.

For question 3, those two circles that you gave don't touch externally. They cut in two places, maybe a typo on your part?

For question 4:

Sin2A = sinA

2sinAcosA = sinA

2sinAcosA - sinA = 0

sinA (2cosA - 1) = 0

sinA = 0, A = 0, 180, 360 (0, pi, 2pi)

2cosA = 1

cosA = 1/2 A = 60, 300 (pi/3, 5pi/3)

shaon0, you lost a couple of solutions

 

[ekoolish]

Oh yeah..

 

[shaon0]

For question 1:

Do what shaon0 said. In addition, find x and y intercepts. y = (x - 1)/(x^2 - 9), Thus when x = 0, y = 1/9 Now when y = 0 , x = 1. The y-intercept is (0, 1/9) and the x - intercept is (1, 0). Now you say you're having problems finding horizontal assymptotes. There is quite a simple way of doing this, which has never failed me. If the denominator has a higher power of x than the numerator, the horizontal assymptote will be y = 0 (in other words, the x - axis). If the denominator has an equal highest power to the numerator, simply cancel these leading coefficients to get your assymptote. For instance: (x^3 + x^2)/(3x^3 + 1), the highest power of numerator and denominator is to the 3rd degree. We cancel the coefficients, and y ----> 1/3, as x ------> infinity. Now the last case is the trickiest type, usually encountered in the extension II course, where the numerator contains a higher power of x than the denominator. An oblique assymptote comes from this situation. In the first two examples we have been given a horizontal assymptote, i.e y = m, m integral. In the last case, you divide top by bottom to get your oblique assymptote. This isn't very important for extension 1, so don't worry too much about it.


For question 2:

If radii add to 100, and one radius is x, therefore the other must be 100-x. Taking both areas:

A1 = pi (x^2)

A2 = pi ((100-x)^2)

= pi (10000 - 200x + x^2)

Now A1 + A2 = pi ( x^2 + 10000 - 200x + x^2) -----> Factoring pi out

= pi (2x^2 - 200x + 10000)

= 2pi (x^2 - 100x + 5000) ----> Factoring 2 out

Now, 2pix^2 - 200pix + 10000pi. Since we want a minimum value, we need to find the vertex.

x value of the vertex is given by -b/2a.

(200pi)/4pi = 50.

Therefore, when x = 50, the minimum area is achieved.

For question 3, those two circles that you gave don't touch externally. They cut in two places, maybe a typo on your part?

For question 4:

Sin2A = sinA

2sinAcosA = sinA

2sinAcosA - sinA = 0

sinA (2cosA - 1) = 0

sinA = 0, A = 0, 180, 360 (0, pi, 2pi)

2cosA = 1

cosA = 1/2 A = 60, 300 (pi/3, 5pi/3)

shaon0, you lost a couple of solutions

Sorry i was busy.

 

[gurmies]

no worries, no need to be sorry

 

[shaon0]

no worries, no need to be sorry

Thanks

 

[bawd]

For question 1:

Do what shaon0 said. In addition, find x and y intercepts. y = (x - 1)/(x^2 - 9), Thus when x = 0, y = 1/9 Now when y = 0 , x = 1. The y-intercept is (0, 1/9) and the x - intercept is (1, 0). Now you say you're having problems finding horizontal assymptotes. There is quite a simple way of doing this, which has never failed me. If the denominator has a higher power of x than the numerator, the horizontal assymptote will be y = 0 (in other words, the x - axis). If the denominator has an equal highest power to the numerator, simply cancel these leading coefficients to get your assymptote. For instance: (x^3 + x^2)/(3x^3 + 1), the highest power of numerator and denominator is to the 3rd degree. We cancel the coefficients, and y ----> 1/3, as x ------> infinity. Now the last case is the trickiest type, usually encountered in the extension II course, where the numerator contains a higher power of x than the denominator. An oblique assymptote comes from this situation. In the first two examples we have been given a horizontal assymptote, i.e y = m, m integral. In the last case, you divide top by bottom to get your oblique assymptote. This isn't very important for extension 1, so don't worry too much about it.


For question 2:

If radii add to 100, and one radius is x, therefore the other must be 100-x. Taking both areas:

A1 = pi (x^2)

A2 = pi ((100-x)^2)

= pi (10000 - 200x + x^2)

Now A1 + A2 = pi ( x^2 + 10000 - 200x + x^2) -----> Factoring pi out

= pi (2x^2 - 200x + 10000)

= 2pi (x^2 - 100x + 5000) ----> Factoring 2 out

Now, 2pix^2 - 200pix + 10000pi. Since we want a minimum value, we need to find the vertex.

x value of the vertex is given by -b/2a.

(200pi)/4pi = 50.

Therefore, when x = 50, the minimum area is achieved.

For question 3, those two circles that you gave don't touch externally. They cut in two places, maybe a typo on your part?

For question 4:

Sin2A = sinA

2sinAcosA = sinA

2sinAcosA - sinA = 0

sinA (2cosA - 1) = 0

sinA = 0, A = 0, 180, 360 (0, pi, 2pi)

2cosA = 1

cosA = 1/2 A = 60, 300 (pi/3, 5pi/3)

shaon0, you lost a couple of solutions

Thanks. Oh and I fixed the question:

3. Prove the circles x^2 + y^2 + 4x - 10y + 20 = 0, x^2 + y^2 - 12x + 2y = 12 touch externally.

And as to shaon0, I didn't get f'(x)=x^2-2x-9 when I differentiated question 1 using quotient rule. I got f'(x)= - x^2 - 2x - 9?