# sol7 - MAE294B/SIO203B Methods in Applied Mechanics Winter...

This preview shows pages 1–2. Sign up to view the full content.

MAE294B/SIO203B: Methods in Applied Mechanics Winter Quarter 2010 http://maecourses.ucsd.edu/mae294b Solution VII 1 The function h ( t ) = - t 4 ( 1 - t ) 2 has maxima at the endpoints 0 and 1 of the interval. The contribution from t = 1 is the standard local maximum times 1 / 2. The contribution from t = 0 is different since h ( 0 ) = 0 and the prefactor t vanishes. One can proceed simply by expanding locally and writing I Z 0 t e - xt 4 d t + Z 1 - e - x ( 1 - t ) 2 d t = 3 4 π x . 2 The function h = - t 3 + 3 t 2 - 2 t is a cubic with a maximum at t 0 = 1 + 1 / 3 in the range of integration. Expanding gives the exact result h ( t ) = h ( t 0 )+ 1 2 h ( t 0 )( t - t 0 ) 2 + 1 6 h ( t 0 )( t - t 0 ) 3 = h ( t 0 ) - 3 ( t - t 0 ) 2 - ( t - t 0 ) 3 . Substitute in: I = Z 0 e x [ h ( t 0 ) - 3 ( t - t 0 ) 2 ] 1 - x ( t - t 0 ) 3 + x 2 2 ( t - t 0 ) 6 + ··· d t . The usual change of variable combined with extending the range gives I e 2 x / 3 3 π 1 / 2 3 - 1 / 4 x - 1 / 2 1 + 5 16 3 x + O ( x - 2 ) . 3 The function h ( x , t ) = xt 2 - t 3 - t 2 / x 2 has potential maxima on the range of integration at 0 or at zeroes of x - 3 t 2 - 2 t / x 2 provided these give maxima. Rather than solving this equation exactly,

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ 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