MATH 1401 #15 QUESTION 2

#15: 2c) ANSWER:

#15: 2d) ANSWER: y = mx + 3 - 2.5m

SOLUTION: The point slope form of a straight line is y = mx + b. The line must go through the point (2.5,3). Therefore, 3 = m(2.5) + b. Solving for b gives b = 3 - 2.5m. Substituting this into the equation results in: y = mx + 3 - 2.5m

#15: 2e) ANSWER: a = - ( 3 - 2.5m ) / m and b = 3 - 2.5m.

SOLUTION: The x-intercept, a, corresponds to y = 0 = ma + 3 - 2.5m. Solving for a results in a = - ( 3 - 2.5m ) / m. The y-intercept is b = 3 - 2.5m.

#15: 2f) ANSWER: s = ( a ² + b ² ) ^{1 / 2}

SOLUTION: The length of the line is the distance between the two points (0,b) and (a,0) which is s = ( a ² + b ² ) ^{1 / 2}

#15: 2g) ANSWER: s = | 3 - 2.5m | ( 1 / m ² + 1 ) ^{1 / 2}

SOLUTION: s = ( a ² + b ² ) ^{1 / 2} = ( (( 3 - 2.5m ) / m) ² + ( 3 - 2.5m ) ² ) ^{1 / 2} = | 3 - 2.5m | ( 1 / m ² + 1 ) ^{1 / 2}

#15: 2h) ANSWER: m < 0

SOLUTION: From the picture, you can see that the slope must be a negative number.

#15: 2i) ANSWER: m = -1.2 ^{1 / 3} ~ -1.063

SOLUTION: NOTE: Since m is negative, | 3 - 2.5m | = 3 - 2.5m . D _m ( | 3 - 2.5m | ( 1 / m ² + 1 ) ^{1 / 2} ) = D _m ( ( 3 - 2.5m ) ( 1 / m ² + 1 ) ^{1 / 2} ) = -2.5 ( 1 / m ² + 1 ) ^{1 / 2} + ( 3 - 2.5m )(1/2)( 1 / m ² + 1 ) ^{- 1 / 2}( -2m ^{- 3} ) = ( 1 / m ² + 1 ) ^{- 1 / 2} ( -2.5 / m ² - 2.5 - 3 / m ³ + 2.5 / m ² ) = - ( 1 / m ² + 1 ) ^{- 1 / 2} ( 2.5 + 3 / m ³ ). This is zero only when 2.5 + 3 / m ³ = 0 or m = - (3/2.5) ^{1 / 3} = -1.2 ^{1 / 3} ~ -1.063.

#15: 2j) ANSWER: m = -1.2 ^{1 / 3} ~ -1.063, s = 2.5 ( 1 + 1.2 ^{2 / 3} ) ^{3 / 2} ~ 7.767

SOLUTION: Since there is only one critical point, there is only one value in the table. When m = -1.2 ^{1 / 3}, s = ( 3 - 2.5( -1.2 ^{1 / 3}) )( 1 / ( -1.2 ^{1 / 3}) ² + 1 ) ^{1 / 2} = 2.5 ( 1 + 1.2 ^{2 / 3} ) ^{3 / 2} ~ 7.767

#15: 2k) ANSWER: From part (j) above, the length is approximately 7.77 feet.