Solutions of the Cambridge Problems: From 1800 to 1820, Volum 2Black, Young, and Young, Tavistock-Street, Covent-Garden., 1825 - 653 sider |
Inni boken
Resultat 1-5 av 58
Side 6
... ratio of BF : FC by n : 1 . Draw FG , Fg parallel to AB , AC respectively , and CN , FM LAB . Then GF : AB :: FC : BF :: 1 : n + 1 AB .. GF = n + 1 Also CN FM :: CB : FB :: n + 1 : n nCN .. FM = also Fg . sin . A n + 1 AB nCN X Hence ...
... ratio of BF : FC by n : 1 . Draw FG , Fg parallel to AB , AC respectively , and CN , FM LAB . Then GF : AB :: FC : BF :: 1 : n + 1 AB .. GF = n + 1 Also CN FM :: CB : FB :: n + 1 : n nCN .. FM = also Fg . sin . A n + 1 AB nCN X Hence ...
Side 14
... ratios between the ordinates which terminate them , we have d . Area ± 2de = = d . log . p 2 = M × de ( M being the modulus of the sys- P tem , and p the radius vector , and the angle described by it ) : .0 = ± 2 Mdp M = CF p2 Let p ...
... ratios between the ordinates which terminate them , we have d . Area ± 2de = = d . log . p 2 = M × de ( M being the modulus of the sys- P tem , and p the radius vector , and the angle described by it ) : .0 = ± 2 Mdp M = CF p2 Let p ...
Side 19
... ratio between the ordinates which intercept it . If CAP = 0 , AP = and CP = x , AC being Ра 1 ( Fig . 15 ) Then z = l . 1 and dz = - P dp P But dz = √dp2 + p2 dė3 , as we readily learn from the figure . :: dp2 + p2 do2 = dp2 : . de ...
... ratio between the ordinates which intercept it . If CAP = 0 , AP = and CP = x , AC being Ра 1 ( Fig . 15 ) Then z = l . 1 and dz = - P dp P But dz = √dp2 + p2 dė3 , as we readily learn from the figure . :: dp2 + p2 do2 = dp2 : . de ...
Side 42
... ratio . Let O be the origin of abscissæ of both the curve and the locus , ON = x ' , CN = y ' , OM = x , PM = y , CD = a , and PD CD = α m m Then , from similar A , we get PD y = y ' x = f ( x ' ) . CD f . ( ON ) a ma m But ON = x - NM ...
... ratio . Let O be the origin of abscissæ of both the curve and the locus , ON = x ' , CN = y ' , OM = x , PM = y , CD = a , and PD CD = α m m Then , from similar A , we get PD y = y ' x = f ( x ' ) . CD f . ( ON ) a ma m But ON = x - NM ...
Side 63
... ratio , viz . 21. Consequently AP is an hyperbola . 59 . Find the equation to a straight line any how disposed in space , and thence deduce the equation to another passing through a given point , at right angles to the former . Let the ...
... ratio , viz . 21. Consequently AP is an hyperbola . 59 . Find the equation to a straight line any how disposed in space , and thence deduce the equation to another passing through a given point , at right angles to the former . Let the ...
Andre utgaver - Vis alle
Solutions of the Cambridge Problems, from 1800 to 1820, Volum 2 John Martin Frederick WRIGHT Uten tilgangsbegrensning - 1836 |
Solutions of the Cambridge Problems: From 1800 to 1820, Volum 2 John Martin Frederick Wright Uten tilgangsbegrensning - 1825 |
Solutions of the Cambridge Problems, from 1800 to 1820, Volum 2 John Martin Frederick Wright Uten tilgangsbegrensning - 1836 |
Vanlige uttrykk og setninger
abscissa accelerating force altitude angular axis b₁ base bisected body centre of gravity chord circle co-declination co-ordinates cone curve cycloid cylinder denote density descending diameter distance dy dx earth ecliptic ellipse equal equation fluid given point gives Hence horizon hyperbola inclination intersection latitude latus rectum length locus logarithmic spiral moving force orbit ordinate orifice oscillation parabola paraboloid parallel perpendicular plane position problem Prop question radius ratio right angles right ascension shew sides specific gravity sphere spherical straight line substituting subtangent supposing surface tangent triangle velocity vers vertex vertical Vince weight whence whole
Populære avsnitt
Side 654 - ... line and the extremities of the base have the same ratio which the other sides of the triangle have to one...
Side 654 - IF a straight line be divided into two equal, and also into two unequal parts; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.
Side 654 - BAC is cut off from the given circle ABC containing an angle equal to the given angle D : Which was to be done. PROP. XXXV. THEOR. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.
Side 654 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.
Side 657 - B. less by 1 1 chains than the length of the sewer ; the expense of making it amounted to as many pounds per chain, as there were chains in the street leading to A. The sewer, however, being insufficient to carry off the water, an additional drain was made from a point in this street distant 4 chains from the bridge A., which entered the river at the same point with the sewer, and was equally inclined to the river and sewer. Now it was found that a drain down the middle of each street, at the rate...
Side 693 - Upon comparing the observations with each other, it was discovered that in both the fore-mentioned stars, the apparent difference of declination from the maxima was always nearly proportional to the versed sine of the sun's distance from the equinoctial points. This was an inducement to think that the cause, whatever it was, had some relation to the sun's situation with respect to those points.
Side 713 - This is the same as saying that when a ray of light passes out of one medium into another, the...
Side 685 - W its weight in water, its weight in vacuo will be, 1 — m 6. Three globes of the same diameter and of given specific gravities, are placed in the same straight line. How must they be disposed that they may balance on the same point of the line in vacuo and in water ? 7. If a homogeneous hemisphere, floating in a fluid, be slightly inclined from the position of equilibrium...
Side 658 - A ship, with a crew of 175 men, set sail with a supply of water sufficient to last to the end of the voyage ; but in 30 days the scurvy made its appearance, and carried off three men every day ; and at the same time a storm arose which protracted the voyage three weeks. They were, however, just enabled to arrive in port without any diminution in each man's daily allowance of water. Required the time of the passage, and the number of men alive when the vessel reached the harbor.
Side 655 - A number of persons purchased a field for £345. The youngest contributed a certain sum, the next £5 more, the third £5 more than the second, and so on to the oldest. For the greater accommodation of the seniors, the field was divided into two parts, the younger half taking a portion proportional to the sum they had subscribed ; and in order that each might have an equal share in this portion, they agreed to equalize their contributions, and each to pay ,£22. Required the number of persons and...