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MA113_2010SPRING_HW1_PROFSOLN_[0]

MA113_2010SPRING_HW1_PROFSOLN_[0] - MA113 Calculus II The...

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MA113 Calculus II The Cooper Union, Spring 2010 Prof. Alex Casti [email protected] Homework 1 Solutions Chapter 9 and 13 problems (Stein and Barcellos) Section 9.3 6 Express the following curve parametrically using the parameter t : y = tan - 1 3 x . Simply choose t = x and then you have x ( t ) = t , y ( t ) = tan - 1 3 t , with t [ a , b ] . The range you choose for t is up to you. 12 Find dy dx and d 2 y dx 2 for the curve defined parametrically by x = e t 2 and y = tan t . First use the fact that dy dx = dy dt dx dt , and then with dy dx expressed as a function of t we use the following expression to get the second derivative: d 2 y dx 2 = d dt dy dx dx dt . Straightforward differentiation gives dx dt = 2 t e t 2 dy dt = sec 2 t dy dx = sec 2 t 2 t e t 2 . The second derivative is then d dt dy dx = - ( 1 + 2 t 2 - 2 t tan t ) sec 2 t 2 t 2 e t 2 d 2 y dx 2 = - ( 1 + 2 t 2 - 2 t tan t ) sec 2 t 4 t 3 e 2 t 2 . 15 Find the equation of the tangent line to the curve [ x ( t ) , y ( t )] = t 3 + t 2 , t 5 + t at the point ( x , y ) = ( 2 , 2 ) . We can use the slope-intercept form for a line, y ( x ) = mx + b , where m is the slope at the given point and b is the y intercept. The point ( 2 , 2 ) corresponds to the parameter value t = 1, so we calculate the slope at this point: m = dy dx t = 1 = dy dt dx dt t = 1 = 5 t 4 + 1 3 t 2 + 2 t t = 1 = 6 5 . The intercept value b can be found by observing y = 2 when x = 2, so 2 = 6 5 ( 2 )+ b b = - 2 5 . 1
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The equation of the line tangent to the curve at t = 1 is thus y = 6 5 x - 2 5 . 18 Find d 2 y dx 2 for the curve given by x ( t ) = e 3 t + sin2 t and y ( t ) = e 3 t + cos t 2 . We use the same expressions as in problem 6, section 9 . 3: dx dt = 3 e 3 t + 2cos2 t dy dt = 3 e 3 t - 2 t sin t 2 dy dx = 3 e 3 t - 2 t sin t 2 3 e 3 t + 2cos2 t . d dt dy dx = ( 2cos2 t + 3 e 3 t )( - 4 t 2 cos t 2 - 2sin t 2 + 9 e 3 t ) + ( 4sin2 t - 9 e 3 t )( 3 e 3 t - 2 t sin t 2 ) ( 3 e 3 t + 2cos2 t ) 2 d 2 y dx 2 = d dt dy dx dx dt = ( 2cos2 t + 3 e 3 t )( - 4 t 2 cos t 2 - 2sin t 2 + 9 e 3 t ) + ( 4sin2 t - 9 e 3 t )( 3 e 3 t - 2 t sin t 2 ) ( 3 e 3 t + 2cos2 t ) 3 . For a fixed value of t , the above expression determines the concavity d 2 y dx 2 at any point on the curve. 19 For which values of t is the following curve concave up and concave down? [ x ( t ) , y ( t )] = t 3 + t + 1 , t 2 + t + 2 Determining the concavity of the curve requires that we find the second derivative d 2 y dx 2 . Proceeding in the usual way we find dx dt = 3 t 2 + 1 dy dt = 2 t + 1 dy dx = 2 t + 1 3 t 2 + 1 . d dt dy dx = 2 - 6 t - 6 t 2 ( 1 + 3 t 2 ) 2 d 2 y dx 2 = d dt dy dx dx dt = 2 - 6 t - 6 t 2 ( 1 + 3 t 2 ) 3 . Now we need to determine the signature of d 2 y dx 2 as a function of t . First, notice that the denominator is positive for all t , so the sign of the second derivative is determined by the numerator 2 - 6 t - 6 t 2 . Setting this quadratic to zero gives two inflection points, which we label as t - and t + : 2 - 6 t - 6 t 2 = 0 t ± = - 1 2 ± 1 6 21 t - - 1 . 26 t + 0 . 26 . Thus, there are three regions of interest on the t axis: ( - , t - ) , ( t - , t + ) , and ( t + , ) . Over these three regions the sign of d 2 y dx 2 remains the same.
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