I need some major help reviewing please

**blackflag10** · November 29th 2011, 05:34 PM

My calculus teacher gave us a review sheet today that has several problems on it that are like what is going to be on our test that we are having Thursday. The problem is that he will be out of town until Thursday morning, so I can't go and ask him for help. I will post two of the problems that I am having issues with here. Please help me!!

1) Consider the polar equation r=24/5+sinx
a)Transform the polar equation into rectangular coordinates.
b) Find the foci of the ellipse formed

2) a) If A is the point (0,0,1) find a point B so that the vector AB has length 6 and is parallel to <-1,4,1>

b) find an equation of the sphere for which the line segment from A to C (4,-6,9) is a diameter

Thanks in advance!!

**Prove It** · November 29th 2011, 06:03 PM

Originally Posted by blackflag10

My calculus teacher gave us a review sheet today that has several problems on it that are like what is going to be on our test that we are having Thursday. The problem is that he will be out of town until Thursday morning, so I can't go and ask him for help. I will post two of the problems that I am having issues with here. Please help me!!

1) Consider the polar equation r=24/5+sinx
a)Transform the polar equation into rectangular coordinates.
b) Find the foci of the ellipse formed

2) a) If A is the point (0,0,1) find a point B so that the vector AB has length 6 and is parallel to <-1,4,1>

b) find an equation of the sphere for which the line segment from A to C (4,-6,9) is a diameter

Thanks in advance!!

Is this $\displaystyle \begin{align*} r = \frac{24}{5} + \sin{x} \end{align*}$ or $\displaystyle \begin{align*} \frac{24}{5\sin{x}} \end{align*}$ ?

**blackflag10** · November 29th 2011, 06:07 PM

It is r= 24/(5+sinx)

**Prove It** · November 29th 2011, 06:20 PM

Originally Posted by blackflag10

It is r= 24/(5+sinx)

You should know that $\displaystyle \begin{align*} x = r\cos{\theta}, y = r\sin{\theta} \end{align*}$ and $\displaystyle \begin{align*} x^2 + y^2 = r^2 \end{align*}$

Therefore

$\displaystyle \begin{align*} r &= \frac{24}{5 + \sin{x}} \\ r &= \frac{24}{5 + \frac{y}{r}} \\ r\left(5 + \frac{y}{r}\right) &= 24 \\ 5r + y &= 24 \\ 5\sqrt{x^2 + y^2} &= 24 - y \\ 25\left(x^2 + y^2\right) &= \left(24 - y\right)^2 \\ 25x^2 + 25y^2 &= y^2 - 48y + 576 \\ 25x^2 + 24y^2 + 48y &= 576 \\ 25x^2 + 24\left(y^2 + 2y\right) &= 576 \\ 25x^2 + 24\left(y^2 + 2y + 1^2 - 1^2\right) &= 576 \\ 25x^2 + 24\left[\left(y + 1\right)^2 - 1\right] &= 576 \\ 25x^2 + 24\left(y + 1\right)^2 - 24 &= 576 \\ 25x^2 + 24\left(y + 1\right)^2 &= 600 \\ \frac{x^2}{24} + \frac{\left(y + 1\right)^2}{25} &= 1 \end{align*}$

You should be able to find the foci now.

**blackflag10** · November 29th 2011, 07:32 PM

Originally Posted by Prove It

You should know that $\displaystyle \begin{align*} x = r\cos{\theta}, y = r\sin{\theta} \end{align*}$ and $\displaystyle \begin{align*} x^2 + y^2 = r^2 \end{align*}$

Therefore

$\displaystyle \begin{align*} r &= \frac{24}{5 + \sin{x}} \\ r &= \frac{24}{5 + \frac{y}{r}} \\ r\left(5 + \frac{y}{r}\right) &= 24 \\ 5r + y &= 24 \\ 5\sqrt{x^2 + y^2} &= 24 - y \\ 25\left(x^2 + y^2\right) &= \left(24 - y\right)^2 \\ 25x^2 + 25y^2 &= y^2 - 48y + 576 \\ 25x^2 + 24y^2 + 48y &= 576 \\ 25x^2 + 24\left(y^2 + 2y\right) &= 576 \\ 25x^2 + 24\left(y^2 + 2y + 1^2 - 1^2\right) &= 576 \\ 25x^2 + 24\left[\left(y + 1\right)^2 - 1\right] &= 576 \\ 25x^2 + 24\left(y + 1\right)^2 - 24 &= 576 \\ 25x^2 + 24\left(y + 1\right)^2 &= 600 \\ \frac{x^2}{24} + \frac{\left(y + 1\right)^2}{25} &= 1 \end{align*}$

You should be able to find the foci now.

Thank you so much!!!

If someone could help me out with the second question, I would be very grateful!

**Prove It** · November 29th 2011, 08:03 PM

Originally Posted by blackflag10

My calculus teacher gave us a review sheet today that has several problems on it that are like what is going to be on our test that we are having Thursday. The problem is that he will be out of town until Thursday morning, so I can't go and ask him for help. I will post two of the problems that I am having issues with here. Please help me!!

1) Consider the polar equation r=24/5+sinx
a)Transform the polar equation into rectangular coordinates.
b) Find the foci of the ellipse formed

2) a) If A is the point (0,0,1) find a point B so that the vector AB has length 6 and is parallel to <-1,4,1>

b) find an equation of the sphere for which the line segment from A to C (4,-6,9) is a diameter

Thanks in advance!!

The vector <-1, 4, 1> represents the vector from the point (0, 0, 0) to (-1, 4, 1).

You wish the vector to be parallel to this and to begin at (0, 0, 1), so to be translated one unit up in the z axis.

So a vector in the direction of <-1, 4, 1> starting at (0, 0, 1) is <-1, 4, 2>.

Now you want this vector to have a length of 6, so if we divide it by its own length, it will be a unit vector, and then multiplying by 6 will give it a length of 6.

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