Harriot, Thomas, Mss. 6787

List of thumbnails

< >
581
581 (291v) 		582
582 (292)
583
583 (292v) 		584
584 (293)
585
585 (293v) 		586
586 (294)
587
587 (294v) 		588
588 (295)
589
589 (295v) 		590
590 (296)
< >
page |< < (313) of 1155 > >|
    <echo version="1.0RC">
      <text xml:lang="eng" type="free">
        <div type="section" level="1" n="1">
          <pb file="add_6787_f313" o="313" n="624"/>
          <div type="page_commentary" level="0" n="0">
            <p>
              <s xml:space="preserve">[
                <emph style="bf">Commentary:</emph>
              </s>
            </p>
            <p>
              <s xml:space="preserve"> This and the pages that follow outline a general method of finding the area of a parabola, based on the fact that sections of the parabola are in constant ratio to their inscribed triangles; and the squares of those triangles increase as the cubes of integers.
                <lb/>
              The calculations on this page are for a parabola whose equation in modern notation is
                <math>
                  <mstyle>
                    <mrow>
                      <msup>
                        <mi>y</mi>
                        <mn>2</mn>
                      </msup>
                    </mrow>
                    <mo>=</mo>
                    <mn>4</mn>
                    <mi>x</mi>
                  </mstyle>
                </math>
              ; thus if the altitude of a given triangle is
                <math>
                  <mstyle>
                    <mi>x</mi>
                  </mstyle>
                </math>
              (the distance along the axis) then the base is
                <math>
                  <mstyle>
                    <mn>2</mn>
                    <mi>y</mi>
                  </mstyle>
                </math>
              (twice the ordinate). </s>
              <s xml:space="preserve">]</s>
            </p>
          </div>
          <head xml:space="preserve"> 1.) The triangles of the parabola. The first base by the </head>
          <p>
            <s xml:space="preserve"> The bases.
              <lb/>
            The altitude.
              <lb/>
            The superficies of the triangle.
              <lb/>
            The superficies of the </s>
          </p>
          <p>
            <s xml:space="preserve"> The progression of the </s>
          </p>
          <p>
            <s xml:space="preserve"> The first triangles base passeth by
              <lb/>
            the </s>
          </p>
          <p>
            <s xml:space="preserve"> The progression of the numerators
              <lb/>
            of the </s>
          </p>
          <p>
            <s xml:space="preserve"> Note.
              <lb/>
            Every triangle is
              <math>
                <mstyle>
                  <mfrac>
                    <mrow>
                      <mn>3</mn>
                    </mrow>
                    <mrow>
                      <mn>4</mn>
                    </mrow>
                  </mfrac>
                </mstyle>
              </math>
            of
              <lb/>
            his parabola.
              <lb/>
            Therefore as:
              <lb/>
            Triangle to triangle
              <lb/>
            so: parabola to parabola:
              <lb/>
              <foreign xml:lang="lat">correspondente</foreign>
            .
              <lb/>
            </s>
          </p>
          <p>
            <s xml:space="preserve"> The halfe triangles: or: parabolæ:
              <lb/>
            or: The rate of the triangles
              <foreign xml:lang="lat">in minimis</foreign>
            </s>
          </p>
          <p>
            <s xml:space="preserve"> That is: the square of the first triangle
              <lb/>
            is the cube of 1.
              <lb/>
            of the second, the cube of 2.
              <lb/>
            of the third, the cube of 3. &</s>
          </p>
        </div>
      </text>
    </echo>
Impressum