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Sagot :
To determine which glider reaches the greater maximum altitude in the first 6 seconds after launch, we need to analyze the altitude functions for both gliders.
Melissa's Altitude Function:
[tex]\[ m(s) = 0.4(s^3 - 11s^2 + 31s - 1) \][/tex]
Robbie's Altitude Function:
[tex]\[ r(s) = 0.4(2s^3 - 7s^2 + 10s + 5) \][/tex]
To find the maximum altitude each glider reaches within the first 6 seconds, we need to perform the following steps for both functions [tex]\( m(s) \)[/tex] and [tex]\( r(s) \)[/tex]:
1. Calculate the first derivative of the function. This will help us find the critical points where the function could have a maximum or minimum.
2. Set the first derivative equal to zero and solve for [tex]\( s \)[/tex]. These are the critical points.
3. Evaluate the function at the critical points and at the endpoints [tex]\( s = 0 \)[/tex] and [tex]\( s = 6 \)[/tex]. This will allow us to find the maximum value within the given interval.
### Step-by-Step Solution for Melissa's Glider:
#### 1. Calculate the first derivative of [tex]\( m(s) \)[/tex]:
[tex]\[ m'(s) = 0.4 \frac{d}{ds}(s^3 - 11s^2 + 31s - 1) \][/tex]
[tex]\[ m'(s) = 0.4(3s^2 - 22s + 31) \][/tex]
[tex]\[ m'(s) = 1.2s^2 - 8.8s + 12.4 \][/tex]
#### 2. Set the first derivative equal to zero and solve for [tex]\( s \)[/tex]:
[tex]\[ 1.2s^2 - 8.8s + 12.4 = 0 \][/tex]
Using the quadratic formula [tex]\( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1.2 \)[/tex], [tex]\( b = -8.8 \)[/tex], and [tex]\( c = 12.4 \)[/tex]:
[tex]\[ s = \frac{8.8 \pm \sqrt{(-8.8)^2 - 4 \cdot 1.2 \cdot 12.4}}{2 \cdot 1.2} \][/tex]
[tex]\[ s = \frac{8.8 \pm \sqrt{77.44 - 59.52}}{2.4} \][/tex]
[tex]\[ s = \frac{8.8 \pm \sqrt{17.92}}{2.4} \][/tex]
[tex]\[ s \approx \frac{8.8 \pm 4.23}{2.4} \][/tex]
[tex]\[ s_1 \approx 5.45 \][/tex]
[tex]\[ s_2 \approx 1.9 \][/tex]
#### 3. Evaluate [tex]\( m(s) \)[/tex] at [tex]\( s = 0 \)[/tex], [tex]\( s = 6 \)[/tex], and the critical points [tex]\( s = 1.9 \)[/tex] and [tex]\( s = 5.45 \)[/tex]:
[tex]\[ m(0) = 0.4(0^3 - 11(0)^2 + 31(0) - 1) = -0.4 \][/tex]
[tex]\[ m(6) = 0.4(6^3 - 11(6)^2 + 31(6) - 1) = 0.4(216 - 396 + 186 - 1) = 0.4(5) = 2 \][/tex]
[tex]\[ m(1.9) = 0.4((1.9)^3 - 11(1.9)^2 + 31(1.9) - 1) \][/tex]
[tex]\[ m(1.9) \approx 0.4(6.859 - 39.71 + 58.9 - 1) \approx 0.4(25.049) = 10.0196 \][/tex]
[tex]\[ m(5.45) = 0.4((5.45)^3 - 11(5.45)^2 + 31(5.45) - 1) \][/tex]
[tex]\[ m(5.45) \approx 0.4(161.17 - 327.32 + 168.95 - 1) \approx 0.4(1.8) = 0.72 \][/tex]
Thus, the maximum altitude for Melissa's glider is approximately [tex]\( 10.0196 \)[/tex] feet.
### Step-by-Step Solution for Robbie's Glider:
#### 1. Calculate the first derivative of [tex]\( r(s) \)[/tex]:
[tex]\[ r'(s) = 0.4 \frac{d}{ds}(2s^3 - 7s^2 + 10s + 5) \][/tex]
[tex]\[ r'(s) = 0.4(6s^2 - 14s + 10) \][/tex]
[tex]\[ r'(s) = 2.4s^2 - 5.6s + 4 \][/tex]
#### 2. Set the first derivative equal to zero and solve for [tex]\( s \)[/tex]:
[tex]\[ 2.4s^2 - 5.6s + 4 = 0 \][/tex]
Using the quadratic formula [tex]\( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2.4 \)[/tex], [tex]\( b = -5.6 \)[/tex], and [tex]\( c = 4 \)[/tex]:
[tex]\[ s = \frac{5.6 \pm \sqrt{(-5.6)^2 - 4 \cdot 2.4 \cdot 4}}{2 \cdot 2.4} \][/tex]
[tex]\[ s = \frac{5.6 \pm \sqrt{31.36 - 38.4}}{4.8} \][/tex]
[tex]\[ s = \frac{5.6 \pm \sqrt{-7.04}}{4.8} \][/tex]
Since the discriminant ([tex]\( -7.04 \)[/tex]) is negative, the quadratic equation has no real roots. Therefore, there are no critical points within the interval [0, 6].
#### 3. Evaluate [tex]\( r(s) \)[/tex] at [tex]\( s = 0 \)[/tex] and [tex]\( s = 6 \)[/tex]:
[tex]\[ r(0) = 0.4(2(0)^3 - 7(0)^2 + 10(0) + 5) = 2 \][/tex]
[tex]\[ r(6) = 0.4(2(6)^3 - 7(6)^2 + 10(6) + 5) = 0.4(432 - 252 + 60 + 5) = 0.4(245) = 98 \][/tex]
Thus, the maximum altitude for Robbie's glider is [tex]\( 98 \)[/tex] feet.
### Conclusion:
Robbie's glider reaches a much greater maximum altitude of [tex]\( 98 \)[/tex] feet compared to Melissa's glider, which reaches a maximum altitude of approximately [tex]\( 10.0196 \)[/tex] feet. Hence,
Robbie's glider reaches the greater maximum altitude in the first 6 seconds after launch.
Melissa's Altitude Function:
[tex]\[ m(s) = 0.4(s^3 - 11s^2 + 31s - 1) \][/tex]
Robbie's Altitude Function:
[tex]\[ r(s) = 0.4(2s^3 - 7s^2 + 10s + 5) \][/tex]
To find the maximum altitude each glider reaches within the first 6 seconds, we need to perform the following steps for both functions [tex]\( m(s) \)[/tex] and [tex]\( r(s) \)[/tex]:
1. Calculate the first derivative of the function. This will help us find the critical points where the function could have a maximum or minimum.
2. Set the first derivative equal to zero and solve for [tex]\( s \)[/tex]. These are the critical points.
3. Evaluate the function at the critical points and at the endpoints [tex]\( s = 0 \)[/tex] and [tex]\( s = 6 \)[/tex]. This will allow us to find the maximum value within the given interval.
### Step-by-Step Solution for Melissa's Glider:
#### 1. Calculate the first derivative of [tex]\( m(s) \)[/tex]:
[tex]\[ m'(s) = 0.4 \frac{d}{ds}(s^3 - 11s^2 + 31s - 1) \][/tex]
[tex]\[ m'(s) = 0.4(3s^2 - 22s + 31) \][/tex]
[tex]\[ m'(s) = 1.2s^2 - 8.8s + 12.4 \][/tex]
#### 2. Set the first derivative equal to zero and solve for [tex]\( s \)[/tex]:
[tex]\[ 1.2s^2 - 8.8s + 12.4 = 0 \][/tex]
Using the quadratic formula [tex]\( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1.2 \)[/tex], [tex]\( b = -8.8 \)[/tex], and [tex]\( c = 12.4 \)[/tex]:
[tex]\[ s = \frac{8.8 \pm \sqrt{(-8.8)^2 - 4 \cdot 1.2 \cdot 12.4}}{2 \cdot 1.2} \][/tex]
[tex]\[ s = \frac{8.8 \pm \sqrt{77.44 - 59.52}}{2.4} \][/tex]
[tex]\[ s = \frac{8.8 \pm \sqrt{17.92}}{2.4} \][/tex]
[tex]\[ s \approx \frac{8.8 \pm 4.23}{2.4} \][/tex]
[tex]\[ s_1 \approx 5.45 \][/tex]
[tex]\[ s_2 \approx 1.9 \][/tex]
#### 3. Evaluate [tex]\( m(s) \)[/tex] at [tex]\( s = 0 \)[/tex], [tex]\( s = 6 \)[/tex], and the critical points [tex]\( s = 1.9 \)[/tex] and [tex]\( s = 5.45 \)[/tex]:
[tex]\[ m(0) = 0.4(0^3 - 11(0)^2 + 31(0) - 1) = -0.4 \][/tex]
[tex]\[ m(6) = 0.4(6^3 - 11(6)^2 + 31(6) - 1) = 0.4(216 - 396 + 186 - 1) = 0.4(5) = 2 \][/tex]
[tex]\[ m(1.9) = 0.4((1.9)^3 - 11(1.9)^2 + 31(1.9) - 1) \][/tex]
[tex]\[ m(1.9) \approx 0.4(6.859 - 39.71 + 58.9 - 1) \approx 0.4(25.049) = 10.0196 \][/tex]
[tex]\[ m(5.45) = 0.4((5.45)^3 - 11(5.45)^2 + 31(5.45) - 1) \][/tex]
[tex]\[ m(5.45) \approx 0.4(161.17 - 327.32 + 168.95 - 1) \approx 0.4(1.8) = 0.72 \][/tex]
Thus, the maximum altitude for Melissa's glider is approximately [tex]\( 10.0196 \)[/tex] feet.
### Step-by-Step Solution for Robbie's Glider:
#### 1. Calculate the first derivative of [tex]\( r(s) \)[/tex]:
[tex]\[ r'(s) = 0.4 \frac{d}{ds}(2s^3 - 7s^2 + 10s + 5) \][/tex]
[tex]\[ r'(s) = 0.4(6s^2 - 14s + 10) \][/tex]
[tex]\[ r'(s) = 2.4s^2 - 5.6s + 4 \][/tex]
#### 2. Set the first derivative equal to zero and solve for [tex]\( s \)[/tex]:
[tex]\[ 2.4s^2 - 5.6s + 4 = 0 \][/tex]
Using the quadratic formula [tex]\( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2.4 \)[/tex], [tex]\( b = -5.6 \)[/tex], and [tex]\( c = 4 \)[/tex]:
[tex]\[ s = \frac{5.6 \pm \sqrt{(-5.6)^2 - 4 \cdot 2.4 \cdot 4}}{2 \cdot 2.4} \][/tex]
[tex]\[ s = \frac{5.6 \pm \sqrt{31.36 - 38.4}}{4.8} \][/tex]
[tex]\[ s = \frac{5.6 \pm \sqrt{-7.04}}{4.8} \][/tex]
Since the discriminant ([tex]\( -7.04 \)[/tex]) is negative, the quadratic equation has no real roots. Therefore, there are no critical points within the interval [0, 6].
#### 3. Evaluate [tex]\( r(s) \)[/tex] at [tex]\( s = 0 \)[/tex] and [tex]\( s = 6 \)[/tex]:
[tex]\[ r(0) = 0.4(2(0)^3 - 7(0)^2 + 10(0) + 5) = 2 \][/tex]
[tex]\[ r(6) = 0.4(2(6)^3 - 7(6)^2 + 10(6) + 5) = 0.4(432 - 252 + 60 + 5) = 0.4(245) = 98 \][/tex]
Thus, the maximum altitude for Robbie's glider is [tex]\( 98 \)[/tex] feet.
### Conclusion:
Robbie's glider reaches a much greater maximum altitude of [tex]\( 98 \)[/tex] feet compared to Melissa's glider, which reaches a maximum altitude of approximately [tex]\( 10.0196 \)[/tex] feet. Hence,
Robbie's glider reaches the greater maximum altitude in the first 6 seconds after launch.
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