PureMath.pdf

5 the constituents of a determinant are functions of

• 587

This preview shows pages 324–327. Sign up to view the full content.

5. The constituents of a determinant are functions of x . Show that its differential coefficient is the sum of the determinants formed by differentiating the constituents of one row only, leaving the rest unaltered. 6. If f 1 , f 2 , f 3 , f 4 are polynomials of degree not greater than 4, then f 1 f 2 f 3 f 4 f 0 1 f 0 2 f 0 3 f 0 4 f 00 1 f 00 2 f 00 3 f 00 4 f 000 1 f 000 2 f 000 3 f 000 4 is also a polynomial of degree not greater than 4. [Differentiate five times, using the result of Ex. 5, and rejecting vanishing determinants.] 7. If y 3 + 3 yx + 2 x 3 = 0 then x 2 (1 + x 3 ) y 00 - 3 2 xy 0 + y = 0. ( Math. Trip. 1903.)

This preview has intentionally blurred sections. Sign up to view the full version.

[VI : 146] DERIVATIVES AND INTEGRALS 309 8. Verify that the differential equation y = φ { ψ ( y 1 ) } + φ { x - ψ ( y 1 ) } , where y 1 is the derivative of y , and ψ is the function inverse to φ 0 , is satisfied by y = φ ( c ) + φ ( x - c ) or by y = 2 φ ( 1 2 x ). 9. Verify that the differential equation y = { x/ψ ( y 1 ) } φ { ψ ( y 1 ) } , where the notation is the same as that of Ex. 8, is satisfied by y = ( x/c ) or by y = βx , where β = φ ( α ) and α is any root of the equation φ ( α ) - αφ 0 ( α ) = 0. 10. If ax + by + c = 0 then y 2 = 0 (suffixes denoting differentiations with re- spect to x ). We may express this by saying that the general differential equation of all straight lines is y 2 = 0. Find the general differential equations of (i) all circles with their centres on the axis of x , (ii) all parabolas with their axes along the axis of x , (iii) all parabolas with their axes parallel to the axis of y , (iv) all circles, (v) all parabolas, (vi) all conics. [The equations are (i) 1 + y 2 1 + yy 2 = 0, (ii) y 2 1 + yy 2 = 0, (iii) y 3 = 0, (iv) (1 + y 2 1 ) y 3 = 3 y 1 y 2 2 , (v) 5 y 2 3 = 3 y 2 y 4 , (vi) 9 y 2 2 y 5 - 45 y 2 y 3 y 4 + 40 y 3 3 = 0. In each case we have only to write down the general equation of the curves in question, and differentiate until we have enough equations to eliminate all the arbitrary constants.] 11. Show that the general differential equations of all parabolas and of all conics are respectively D 2 x ( y - 2 / 3 2 ) = 0 , D 3 x ( y - 2 / 3 2 ) = 0 . [The equation of a conic may be put in the form y = ax + b ± p px 2 + 2 qx + r. From this we deduce y 2 = ± ( pr - q 2 ) / ( px 2 + 2 qx + r ) 3 / 2 . If the conic is a parabola then p = 0.] 12. Denoting dy dx , 1 2! d 2 y dx 2 , 1 3! d 3 y dx 3 , 1 4! d 4 y dx 4 , . . . by t , a , b , c , . . . and dx dy , 1 2! d 2 x dy 2 , 1 3! d 3 x dy 3 , 1 4! d 4 x dy 4 , . . . by τ , α , β , γ , . . . , show that 4 ac - 5 b 2 = (4 αγ - 5 β 2 ) 8 , bt - a 2 = - ( βτ - α 2 ) 6 .
[VI : 146] DERIVATIVES AND INTEGRALS 310 Establish similar formulae for the functions a 2 d - 3 abc - 2 b 3 , (1 + t 2 ) b - 2 a 2 t , 2 ct - 5 ab . 13. Prove that, if y k is the k th derivative of y = sin( n arc sin x ), then (1 - x 2 ) y k +2 - (2 k + 1) xy k +1 + ( n 2 - k 2 ) y k = 0 . [Prove first when k = 0, and differentiate k times by Leibniz’ Theorem.] 14. Prove the formula vD n x u = D n x ( uv ) - nD n - 1 x ( uD x v ) + n ( n - 1) 1 · 2 D n - 2 x ( uD 2 x v ) - . . . where n is any positive integer. [Use the method of induction.] 15. A curve is given by x = a (2 cos t + cos 2 t ) , y = a (2 sin t - sin 2 t ) .

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.
• Fall '14

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern