I have a problem. I am currently working on my Calculus homework and my professor has the following problem: Graph the function f(x) = (such and such) using the scheme for graphing. What the hell is "the scheme for graphing?" I've never even heard that terminology before. A million internets to the /b/tard that helps me out.
Name:
Anonymous2006-11-27 0:26
If you have a book, it's in there somewhere.
My guess is that it's that thing where you use a functions derivatives to find out where/how the graph bends and such.
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
=\frac{880000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \sqrt{1-\sin^2 \theta} (-90 \cos \theta) d\theta =-\frac{7920000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \cos^2 \theta d\theta =-\frac{7920000}{231}\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} (we did the integral of \cos^2 \theta in problem 1)
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
=\frac{880000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \sqrt{1-\sin^2 \theta} (-90 \cos \theta) d\theta =-\frac{7920000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \cos^2 \theta d\theta =-\frac{7920000}{231}\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} (we did the integral of \cos^2 \theta in problem 1) =-\frac{7920000}{231} \left[\frac{\sin^{-1}\left(1-\frac{h}{90}\right)}{2} - \frac{\pi}{4} + 2 \sin \sin^{-1}\left(1-\frac{h}{90}\right) \cos \sin^{-1}\left(1-\frac{h}{90}\right)\right] =-\frac{7920000}{231} \left[\frac{\sin^{-1}\left(1-\frac{h}{90}\right)}{2} - \frac{\pi}{4} + 2 \left(1-\frac{h}{90}\right) \sqrt{1-\left(1-\frac{h}{90}\right)^2}\right]
Problem 5)
Here's the picture:
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
=\frac{880000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \sqrt{1-\sin^2 \theta} (-90 \cos \theta) d\theta =-\frac{7920000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \cos^2 \theta d\theta =-\frac{7920000}{231}\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} (we did the integral of \cos^2 \theta in problem 1) =-\frac{7920000}{231} \left[\frac{\sin^{-1}\left(1-\frac{h}{90}\right)}{2} - \frac{\pi}{4} + 2 \sin \sin^{-1}\left(1-\frac{h}{90}\right) \cos \sin^{-1}\left(1-\frac{h}{90}\right)\right] =-\frac{7920000}{231} \left[\frac{\sin^{-1}\left(1-\frac{h}{90}\right)}{2} - \frac{\pi}{4} + 2 \left(1-\frac{h}{90}\right) \sqrt{1-\left(1-\frac{h}{90}\right)^2}\right]
Problem 5)
Here's the picture:
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
=\frac{880000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \sqrt{1-\sin^2 \theta} (-90 \cos \theta) d\theta =-\frac{7920000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \cos^2 \theta d\theta =-\frac{7920000}{231}\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} (we did the integral of \cos^2 \theta in problem 1)
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
=\frac{880000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \sqrt{1-\sin^2 \theta} (-90 \cos \theta) d\theta =-\frac{7920000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \cos^2 \theta d\theta =-\frac{7920000}{231}\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} (we did the integral of \cos^2 \theta in problem 1) =-\frac{7920000}{231} \left[\frac{\sin^{-1}\left(1-\frac{h}{90}\right)}{2} - \frac{\pi}{4} + 2 \sin \sin^{-1}\left(1-\frac{h}{90}\right) \cos \sin^{-1}\left(1-\frac{h}{90}\right)\right] =-\frac{7920000}{231} \left[\frac{\sin^{-1}\left(1-\frac{h}{90}\right)}{2} - \frac{\pi}{4} + 2 \left(1-\frac{h}{90}\right) \sqrt{1-\left(1-\frac{h}{90}\right)^2}\right]
Problem 5)
Here's the picture:
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
=\frac{880000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \sqrt{1-\sin^2 \theta} (-90 \cos \theta) d\theta =-\frac{7920000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \cos^2 \theta d\theta =-\frac{7920000}{231}\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} (we did the integral of \cos^2 \theta in problem 1)
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
=\frac{880000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \sqrt{1-\sin^2 \theta} (-90 \cos \theta) d\theta =-\frac{7920000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \cos^2 \theta d\theta =-\frac{7920000}{231}\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} (we did the integral of \cos^2 \theta in problem 1) =-\frac{7920000}{231} \left[\frac{\sin^{-1}\left(1-\frac{h}{90}\right)}{2} - \frac{\pi}{4} + 2 \sin \sin^{-1}\left(1-\frac{h}{90}\right) \cos \sin^{-1}\left(1-\frac{h}{90}\right)\right] =-\frac{7920000}{231} \left[\frac{\sin^{-1}\left(1-\frac{h}{90}\right)}{2} - \frac{\pi}{4} + 2 \left(1-\frac{h}{90}\right) \sqrt{1-\left(1-\frac{h}{90}\right)^2}\right]
Problem 5)
Here's the picture:
We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated. Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture. If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes. If not, there's always wikipedia. Anyway, the equation of this ellipse turns out to be
Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices. Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation
and the length of the line we want to integrate is twice this. Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes
\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy =\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
=\frac{880000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \sqrt{1-\sin^2 \theta} (-90 \cos \theta) d\theta =-\frac{7920000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \cos^2 \theta d\theta =-\frac{7920000}{231}\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} (we did the integral of \cos^2 \theta in problem 1)