A square is inscribed in a circle. As the square expands, the circle expands to maintain the four points of intersection. The perimeter of the square is expanding at the rate of 8 inches per second.
Find the rate at which the circumference of the circle is increasing.
Perimeter = p...
Oh, I figured it out. Turns out that 141.7 = a . So then using the eccentricity I solved for c and then for b using the relationship between the 3. Then the max is a + c and the min is b - c. :)
OK, so
.093 = 1 - \frac{b}{a}
and
.093^2 = 1 - \frac{b^2}{a^2}
But if I rewrite the first I get a = .093a+b That can't be right because I still have 2 variables.
Um, a^2 = b^2 + c^2 is the same thing as the formula I posted...
But, either way, yout idea is not right. 141.7 million is the mean distance not the distance to any point on the orbit. I think the sun is supposed to be at the focus.
Having trouble with this problem.
"The mean distance from the sun to Mars is 141.7 million miles. If the eccentricity of the orbit of Mars is .093, determine the maximum distance that Mars orbits from the sun."
So basically what it is asking for is half the length of the major axis right...
I don't want coordinates for when the value equals 0. All I want, is to convert the rectangular equation into polar form. So instead of having x and y I need r and \Theta.
Currently, I have r^2-3cos\Theta+4sin\Theta=0 but I don't know how to convert the sin and cos into polar form.
Well, actually, I guess they are the same thing, but either way, I don't know how to do it :(
As for your question, I don't see a purpose, but my teacher sure does.
I need to convert x^2+y^2-3cos\Theta+4sin\Theta=0 to polar.
Obviously the x^2+y^2 part would = r^2, but how can I get the cos and sin part to simplify?
I have to find (a+bi)(c+di) in polar form given that b,c,d>0 and a<0.
So I convert each one to polar first.
( (a+b)cis(\arctan(-b/a) + \pi) ) ( (c+d)cis(\arctan(d/c)) )
That's as far as I got. Little help please?
Yeah, I figured out the tan(x) thing. These were due today. I forget how I did the 2nd one because I didn't do it like you said. It worked some other way. What is the purpose of learning precal when the calculator already told me the answer :p
Um HallsofIvy how was the 3 seemingly unaffected by you multiplying through by sin and cos? And when I type in sin^-1(+/- root 5 / 2) I get domain error.... ?
What do you mean by let y=sin(x)?
Oh, my teacher told me sin and cos usually mess things up, but I guess that is the only option.
Edit:
I ended up with 2sin(x)^2 - 2cos(x)^2 + 3cox(x)sin(x) = 0 for the first one. That can't be right can it?
And using the identity on the 2nd one I get:
-sin(x)^2 + sin(x) +1 = 0
Then what...
I'm having trouble with two problems:
2tan(x) - 2cot(x) = -3
and
cos(x)^2 + sin(x) = 0
On the 2nd one, I can substitute 1-sin(x)^2 for cos(x)^2 right? I tried that, but it didn't work. And I have no clue what to do on the first one. Little help please?
OK, thanks.
On the 2nd part, it has nothing to do with graphing. There is just a graph with f and g drawn on it and you just have to find the value based on that graph. All the other questions were normal but I don't understand the whole at x = whatever business. That why I asked if it was...
Little help please and some work checking.
Given f(x)=3x+2 and g(x)=\frac{x-4}{2x}
It asks for g(\frac{1}{x})
So substitute in: \frac{\frac{1}{x} - 4}{2(\frac{1}{x})}
Simplify: \frac{\frac{1}{x} - 4}{\frac{2}{x}}
I multiply top and bottom by x right?
That would give: \frac{1 -...
\frac{x}{x^2-9} - \frac{1}{2x-6}
When I first worked this problem I found the answer to be:
\frac{1}{2x-6}
However, in my English class we had this Vietnamese substitute who took my worksheet and did the problems on it in his head and pointed at the answer I wrote for this problem and...
I havn't seen the one you're talking about with the pyroclastic car chase. (o_0 ?) I saw one like last month on the Discovery Channel but I found that it was actually made by the BBC back in around 2001 or something, so I guess they have been making other ones.
You have to get it in standard form first. Divide both sides by 144.
That gives:
\frac{x^2}{16} + \frac{y^2}{9} = 1
a = length of the major axis
b= length of the minor axis