sol_diagnostic

sol_diagnostic - θ with 0 ≤ θ< 2 π with sin θ =-√...

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KEY Diagnostic 1. Find the center of the circle whose equation is x 2 + y 2 = - 4 x + 6 y x 2 + 4 x + y 2 - 6 y = 0 x 2 + 4 x + 4 + y 2 - 6 y + 9 = 13 ( x + 2) 2 + ( y - 3) 2 = 13 ( x + 2) 2 + ( y - 3) 2 = ( 13) 2 The center is ( - 2 , 3) and the radius is 13. 2. Find the equation of the line containing the two points ( - 1 , 3) and (2 , 4). slope = m = 4 - 3 2 - ( - 1) = 1 3 . We have y = m * x + b and y = 1 3 x + b . Now, we need to solve for b . We will use the point (2 , 4) so 4 = 1 3 · 2 + b implies b = 10 3 and y = 1 3 x + 10 3 . 3. log 8 (4) = log 8 (8 2 / 3 ) = 2 / 3 4. (2 2 ) 8 = 2 16 = 2 4 = 16 5. Solve the inequality | 2 x - 1 | < 1 / 3. Write your answer in interval notation. - 1 / 3 < 2 x - 1 < 1 / 3 2 / 3 < 2 x < 4 / 3 1 / 3 < x < 2 / 3 In interval notation we have (1 / 3 , 2 / 3). 6. If θ is in the second quadrant and cos( θ ) = - 3 / 5, ﬁnd tan( θ ). Recall: sin 2 ( θ )+cos 2 ( θ ) = 1. 9 / 25 + sin 2 ( θ ) = 1 implies sin 2 ( θ ) = 16 / 25. In the second quadrant sine is positive, so sin( θ ) = 4 / 5. Since tan( θ ) = sin( θ ) / cos( θ ) = - 4 / 3. (You may also use SOHCATOA if you know what that means.) 7. Find all
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Unformatted text preview: θ with 0 ≤ θ < 2 π with sin( θ ) =-√ 3 / 2. Recall: sin( π/ 3) = √ 3 / 2 and sine is negative in quadrant III and IV. So 2 π-π/ 3 = 5 π/ 3 and π + π/ 3 = 4 π/ 3 8. Find lim x → sin( x 2 ) x 2 lim x → sin( x 2 ) x 2 L = lim x → 2 x cos( x 2 ) 2 x = cos(0) = 1 *Note that L = means that L ’hospital’s theorem was invoked. 9. If f ( x ) = 3-x , ﬁnd f 00 ( x ). Let y = 3-x so ln( y ) =-x ln(3) and y y =-ln (3) which imples that y =-ln(3)3-x . So y 00 = d dx y =-ln(3) d dx 3-x = (ln(3)) 2 3-x . 10. Compute R cos(ln( x )) x dx. Let u = ln x and du = 1 x dx . Thus, Z cos(ln( x )) x dx = Z cos ( u ) du = sin( u ) + c = sin(ln( x )) + c...
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