SalasSV_07_09_ex

# A x c a is a solution of the equation which satises v0

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: es 41–43, verify the given integration formula. In each case, take a > 0. 41. 42. √ √ 1 a2 + x2 1 x 2 − a2 dx = sinh−1 dx = cosh −1 x + C. a x + C. a is a solution of the equation which satisﬁes v(0) = 0. (b) Find lim v(t ). t →∞ This limit is called the terminal velocity of the body. c 45. Use a CAS to ﬁnd a one-term expression for f . (a) (b) f (x) dx = − f (x) dx = sech x(2 + tanh x) . 3(1 + tanh x)2 1 tanh−1 x + C If |x| < a. a 1 a 43. dx = 1 a2 − x 2 coth−1 x + C If |x| > a. a a 44. If a body of mass m falling from rest under the action of gravity encounters air resistance that is proportional to the square of its velocity, then the velocity v(t ) of the body at time t satisﬁes the equation m dv = mg − kv2 dt ln (x4 − 1) x2 tanh−1 x2 + . 2 4 Verify your results by differentiating the right-hand sides. c 46. Use a CAS to ﬁnd a one-term expression for f . x x (a) f (x) dx = 16 + x2 − 8 sinh−1 . 2 4 x2 + 1 x2 −1 x2 cosh−1 x2 + . 2 x2 + 1 2 Verify your results by differentiating the right-hand sides. (b) f (x) dx = − where k > 0 is the constant of proportionality and g is the gravitational constant. CHAPTER HIGHLIGHTS 7.1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online