Kuroyuki's Spesh thread

[Kuroyuki]

The question is find the area in the graph (x^2)/9  +  (y^2)  = 1
The answers did something like dx/dtheta could someone explain that?
Thanks in advance,
Kuroyuki

[brightsky]

The area is simply pi*3*1 = 3pi units^2. You know how the area of a circle is given by pi*r^2? Well, the area of an ellipse is given by pi*a*b, where a and b are the semi-major and semi-minor axes of the ellipse respectively.

I'm not sure what exactly the textbook did, but you can find the area using calculus by integrating sqrt(1-(x^2/9)) with respect to x from x = 0 to x = 3 and then multiply the answer by 4.

[Kuroyuki]

The area is simply pi*3*1 = 3pi units^2. You know how the area of a circle is given by pi*r^2? Well, the area of an ellipse is given by pi*a*b, where a and b are the semi-major and semi-minor axes of the ellipse respectively.

I'm not sure what exactly the textbook did, but you can find the area using calculus by integrating sqrt(1-(x^2/9)) with respect to x from x = 0 to x = 3 and then multiply the answer by 4.

oh thanks!
The book integrated that by letting x = 3 sin theta and then made the integral with respect to theta and evaluated it.

[brightsky]

Oh yes that is usually how you integrate stuff in the form sqrt(a^2 - x^2). You perform what is called a trig substitution, and use the property that sin^2(x) + cos^2(x) = 1. See http://en.wikipedia.org/wiki/Trigonometric_substitution for further details.

[Kuroyuki]

Help with
A solid sphere of radius 6cm has a cylindrical hole of radius 1 cm bored through its centre. What is the volume of the remainder of the sphere?
Thanks in advance

[brightsky]

Volume = 4/3*pi*(6)^3 - pi*1*12

[Kuroyuki]

Volume = 4/3*pi*(6)^3 - pi*1*12

Oh lol thank you,  the worked solutions used calculus and had a page worth of working. 

[abcdqd]

From what I remember, it isn't that simple - the "cylindrical hole" isn't just a cylinder carved out of the sphere. Rather, imagine a hole being drilled through the center of the sphere. Due to the curvature of the sphere, a sort of "dome" will be cut off from both ends of the sphere along with the typical cylinder shape. 

[Kuroyuki]

Volume = 4/3*pi*(6)^3 - pi*1*12

Actually that gives a different answer to the one in the book and the worked solutions. The answer given is (140*(sqrt35) * pi)/3. I think it might have to do with the sphere not having a flat top.
EDIT -

From what I remember, it isn't that simple - the "cylindrical hole" isn't just a cylinder carved out of the sphere. Rather, imagine a hole being drilled through the center of the sphere. Due to the curvature of the sphere, a sort of "dome" will also be cut off from both ends of the sphere with the typical cylinder shape. 

Thanks, ill try look at the worked solutions and make sense of it.

[Conic]

A graph of (i.e. a circle of radius 6):

Spoiler

You can see the top of the volume removed isn't flat, so you need to find the volume of the curved "caps" on the cylinder removed.
To find the volumes of the caps, we first need to find the y coordinates of the points of intersection:



To get the volume of the cap, we revolve the area bound by and :



Since there are 2 caps, the volume is

Now we need to find the volume of the cylinder. The height of the cylinder is 2 times the y coordinate, i.e. .



Now we have the volume of the caps, and the cylinder. The total volume removed is the sum of the caps and the cylinders.



To find the remaining volume, we need to find the original volume:



We have the original volume and the volume removed, so now we can find the new volume:

[Kuroyuki]

I understand now, thank so much!

[Conic]

Actually, there's a much quicker way of doing it. Just revolve the area bound by x=1 and the circle around the y axis:

[Kuroyuki]

Haven't used this thread in a while.
Could someone please help me with
'On an argand diagram A and B represent the complex numbers a= 5 + 2i and b = 8 + 6i
a) Find i(a-b) and show that it can be represented by a vector perpendicular to AB and of the same length as AB.
b) Hence, find complex numbers c and d represented by c and d such that ABCD is a square."
I can do part a just not part b.
Thanks in advance!

[d3stiny]

Notice that when you multiply a-b by i, you rotate the complex number by 90 degrees anticlockwise about the origin.

This gives you another side of the square since each side of a square is 90 deg apart.
However the i(a-b) is relative to the origin, so add it to the point A to make it start from A, since complex numbers can be added like vectors (or B depending where you want the side to join up).

Multiply a-b by i again and you will get another side and so on until the square is complete.

[Kuroyuki]

Thank you!!

So the question is
A curve has parametric equations x = t - cos t  and y = sin t
find the tangent to the curve when t = pi/6
If I use dy/dx = dy/dt * dt/dx i get the correct gradient of root3/3 and a tangent to the curve.
But if i try use the sin^2 t + cos^2 t = 1 identity then implicitly differentiate i get my gradient of the tangent to be root3. And from my calculations doesn't give a tangent to the curve.
Am I differentiating wrong or is implicit differentiation inappropriate or something for this question?
Thanks in advance!