{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MA113_2009SPRING_EXAM1_PROFSOLN_[0]

# MA113_2009SPRING_EXAM1_PROFSOLN_[0] - MA113 Calculus II The...

This preview shows pages 1–3. Sign up to view the full content.

MA113 Calculus II The Cooper Union, Spring 2009 Prof. Alex Casti Midterm Exam Solutions 2 hours Problem 1 (25 points) Give brief answers to the following questions, which may involve short mathematical calcula- tions. (a) The acceleration a ( t ) for a trajectory can be written as a weighted sum of a normal and tangential component. Which of these components do you expect to dominate on the parts of the path with high curvature? Why? The normal component of acceleration should dominate in regions of high curvature. This intuition comes from the fact that the acceleration vector can be written as a ( t ) = a T ˆ T + a N ˆ N , where a T = d 2 s dt 2 is the acceleration tangential to the path and a N = κ v 2 is the normal component of acceleration. Here s ( t ) is arc length, t the parameter of the orbit, v = ds dt the speed, and κ = ± ± ± d φ ds ± ± ± the curvature. In general, if κ v 2 ± d 2 s dt 2 in any particular region due to large κ , then the normal acceleration a N ˆ N will dominate. This could even be the case if the curvature is not particularly large, but the speed v = ds dt is constant, in which case a T = 0. (b) Consider the parametric curve described by x ( t ) = 1 + ln t and y ( t ) = t ln t . Calculate dy dx and d 2 y dx 2 , but do not begin by writing y = y ( x ) to do it. For this calculation we need the formulas dy dx = dy dt dx dt d 2 y dx 2 = d dt ² dy dx ³ dx dt . Doing the straightforward differentiations gives dx dt = 1 t dy dt = 1 + ln t dy dx = t ( 1 + ln t ) d dt ´ dy dx µ = 2 + ln t d 2 y dx 2 = t ( 2 + ln t ) . (c) Describe how you can calculate the volume of a sphere with the disc method by revolving a semi-circle around an axis. Set up the appropriate integral, but you do not need to evaluate the integral. In the disc method, ﬁrst draw the appropriate curve y = y ( x ) over a domain x [ a , b ] and revolve it around a convenient axis (say the x -axis). This will form a solid of revolution. Each inﬁnitesimal volume element dV is approximated by the inﬁnitesimal thickness dx times a cross-sectional area A ( x ) = π y 2 ( x ) , since y corresponds to the radius of the disc formed by revolving the curve. We thus have dV = π y 2 ( x ) dx and the volume of the solid of revolution is V = R b a π y 2 ( x ) dx . In this case, choose the upper branch of the circle of radius a : y ( x ) = a 2 - x 2 for x [ - a , a ] . The volume of the sphere is then given by V = π Z a - a ² p a 2 - x 2 ³ 2 dx 1

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

View Full Document
= π Z a - a ( a 2 - x 2 ) dx = 2 π Z a 0 ( a 2 - x 2 ) dx = 2 π ± a 2 x - 1 3 x 3 ² a x = 0 = 2 π ³ a 3 - 1 3 a 3 ´ V = 4 π a 3 3 . (d) Prove that a time-dependent vector r
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

MA113_2009SPRING_EXAM1_PROFSOLN_[0] - MA113 Calculus II The...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online