Download Module 8 Lesson 4: Applications of Vectors

Module 8 Lesson 4: Applications of Vectors So now that you have learned the basic skills necessary to understand and operate with vectors, in this lesson, we will look at how to solve real world problems that involve vectors. As stated earlier, vectors often involve applications of force, work, weight, navigation, and a few other topics. Let’s go ahead and get started. It will be best if you have your Notes from Lessons 1-­‐3 handy in case you need to reference them. Using Vectors to Find Speed & Direction Ex1. An airplane is flying on a bearing of 341° at 560 mph. Find the component form of the velocity of the plane. Solution: It helps to sketch a picture of the situation. If v has direction angle 𝜃, the components of v can be computed using the formula below. 𝐯 = 𝐯 cos 𝜃 , 𝐯 sin 𝜃 Determine the magnitude of v. Here in this instance, if the plane is flying at 560 mph, then the magnitude of the plane (the length that it flies) will be 560 miles (per hour in flight). 𝐯 = 560 To determine the direction angle, recall that this is the counterclockwise angle between the vector v and the positive x-­‐axis. The problem statement gives the 1 bearing of v, which is the measure of the clockwise angle between v and the y-­‐
axis. The following graph correctly demonstrates the direction of v. Notice that the measure of the clockwise angle formed by the vector and the y-­‐axis is the bearing 341°. We can determine the direction angle of v. 𝜃 = 90° + 360° − 341° = 109° Now, we compute the component form of the velocity of the angle. 𝐯 = 𝐯 cos 𝜃 , 𝐯 sin 𝜃 = 560 cos 109° , 560 sin 109°
= −182.318, 529.490 2 Ex2. An airplane is flying on a compass heading (bearing) of 350° at 355 mph. A wind is blowing with the bearing 310° at 60 mph. (a) Find the component form of the velocity of the plane. (b) Find the actual ground speed and direction of the plane. Solution for (a): Find vectors a and b that model the velocity of the airplane and the velocity of the wind, then use these vectors to determine the ground speed and bearing of the plane. Remember that the bearing is the clockwise angle of the velocity with the positive y-­‐axis. If a vector v has direction angle 𝜃, the components of v can be computed using the following formula. 𝐯 = 𝐯 cos 𝜃 , 𝐯 sin 𝜃 The bearing of the airplane is 350°. Let 𝛼 be the direction angle of the airplane’s velocity without wind. Determine the measure of 𝛼. 𝛼 = 100° Compute the components of a, the vector of the airplane’s velocity without wind. 𝐚 = 𝐚 cos 𝛼 , 𝐚 sin 𝛼 = 335 cos 100° , 335 sin 100°
= −61.645, 349.607 The bearing of the wind is 310°. Let 𝛽 be the direction angle of the wind. Determine the measure of 𝛽 . 3 𝛽 = 140° Compute the component form of b, the velocity of the wind. 𝐛 = 𝐛 cos 𝛽 , 𝐛 sin 𝛽 = 60 cos 140° , 60 sin 140° = −45.963, 38.567 The true velocity of the plane is 𝐯 = 𝐚 + 𝐛. Add these vectors componentwise to find v. 𝐯 = 𝐚 + 𝐛 = −61.645, 349.607 + −45.963, 38.567
= −107.608, 388.174 Solution for (b): The ground speed of the airplane is the magnitude of the velocity of the airplane. Recall that if, 𝐯 = a, b , then 𝐯 = 𝑎! + 𝑏 ! . Compute the ground speed of the airplane. 𝐯 ≈ −107.608 𝟐 + 388.174 𝟐 ≈ 402.813 The ground speed of the airplane is approximately 402.813 mph. Now, all that is left to find is the bearing for the ground speed of the airplane. From the formula for writing a vector in component form, we know the first component of the velocity vector (which we found to be −107.608) can be written as 𝐯 cos 𝜃, where 𝜃 is the directional angle. We can set up the equation 402.813 ∙ cos 𝜃 = −107.608 4 and solve for 𝜃. −107.608
cos 𝜃 =
402.813
𝜃 = cos !!
−107.608
402.813
Compute 𝜃. Note that since the x-­‐component of v is negative and its y-­‐
component is positive, this angle should occur in Quadrant II. 𝜃 = 105.494° Determine the bearing to which 105° corresponds. The bearing = 360° − 105.494° − 90° = 345.506° Therefore, the ground speed is approximately 403 mph with bearing 346°. Ex3. A basketball is shot at a 60° angle with the horizontal direction with an initial speed of 42 feet per second. Find the component form of the initial velocity. Solution: Let v be a nonzero vector. If 𝜃 is the direction angle measured from the positive x-­‐axis to v, then the vector can be expressed in terms of its magnitude and direction angle by the formula below. 𝐯 = 𝐯 cos 𝜃 , 𝐯 sin 𝜃 5 Let’s find the value of the direction angle measured from the positive x-­‐axis to the vector v. 𝜃 = 60° 6 Since the basketball is released with a speed of 42 feet per second, the magnitude of v is 42 feet per second. 𝐯 = 42 Now, substitute the values for 𝐯 = 42 and 𝜃 = 60° into our formula to express v in terms of its magnitude and directional angle. 𝐯 = 𝐯 cos 𝜃 , 𝐯 sin 𝜃 = 42 cos 60° , 42 sin 60° = 21, 36.373 Thus, the vector v is expressed in component form 21, 36.373 . Finding a Force The previous examples are applications in which two vectors are added to produce a resultant vector. Many applications in physics and engineering pose the reverse problem – decomposing a given vector into the sum of two vector components. Consider a boat being pulled on an inclined ramp (shown above). The force F due to gravity pulls the boat down and against the ramp. These two orthogonal (perpendicular) vectors, 𝐰! + 𝐰! , are vector components of F. 𝐅 = 𝐰! + 𝐰! The negative of component 𝐰! represents the force needed to keep the boat from rolling down the ramp, and 𝐰! represents the force that the tires must withstand against the ramp. 7 To find the force required in this problem, we will use the formula 𝐅∙𝐯
proj𝐯 𝐅 =
𝐯 𝐯 !
Ex4. A 600 pound boat sits on a ramp inclined at 30°. What force is required to keep the boat from rolling down the hill? (In other words, we are only looking for 𝐰! .) 8 Solution: Because the force due to gravity is vertical and downward and has a magnitude equal to the combined weight, you can represent the gravitational force by the vector 𝐅 = −600𝐣 The ramp is inclined at 30°. To find the force required to keep the boat from rolling down the ramp, we can project F onto a unit vector v in the direction of the ramp, as follows. 3
1
𝐯 = cos 30° 𝐢 + sin 30° 𝐣 =
𝐢 + 𝐣 2
2
Therefore the projection of F onto v is as follows. Remember Note: In this example, that 𝐯 = 1 if it v is a unit vector. we will refrain from using our calculators and therefore practice our “unit circle” knowledge to evaluate our trigonometric functions. 𝐰! = proj𝐯 𝐅 =
𝐅∙𝐯
𝐯 = 𝐅 ∙ 𝐯 𝐯 𝐯 !
It’s easiest to find (𝐅 ∙ 𝐯) first using our Dot Product knowledge. 3
1
𝐅 ∙ 𝐯 = −600𝐣 ∙
𝐢 + 𝐣 2
2
= 0∙
3
1
+ −600 ∙ 2
2
= 0 − 300 = −300 So now we can substitute this value 𝐅 ∙ 𝐯 𝐯 = −300
3
1
𝐢 + 𝐣 2
2
The magnitude of this force is 300, and therefore a force of 300 pounds is required to keep the boat from rolling down the hill. Finding Work The work W done by a force F as its point of application moves along the vector 𝑃𝑄 is given by 𝑾 = 𝐅 ∙ 𝑷𝑸 Ex5. Find the work done by a force F of 12 pounds acting in the direction 2,2 in moving an object 7 feet from (0, 0) to (7, 0). Solution: The work done by a force F moving an object from P to Q is 𝑾 = 𝐅 ∙ 𝑃𝑄. If F has a magnitude 12 and acts in the direction 2,2 , then the following is true 2,2
12
12
12
𝐅 = 12
=
2,2 =
2,2 =
2,2 2,2
2,2
2 2
2! + 2!
Since the object is being moved from point (0, 0) to (7, 0), then PQ = 7,0 . 𝐖 = 𝐅 ∙ 𝐏𝐐 9 =
12
2 2
2,2
7,0 Use the Dot Product to find 2,2 ∙ 7,0 . 2,2 ∙ 7,0 = 2 ∙ 7 + 2 ∙ 0 = 14 + 0 = 14 Therefore, we have !"
∙ 14 ≈ 59.397 foot-­‐pounds. ! !
Geometric Applications: Area For applications of 3D vectors, we will be calculating the area of figures in the 3D plane. Here in this example, we will find the area of a parallelogram. Ex6. Find the area of a parallelogram with vertices 𝐴 5, 2, 0 , 𝐵 2, 6, 1 , 𝐶 2, 4, 7 , and 𝐷 5, 0, 6 . Solution: Let’s first sketch these points in the xyz plane. The area formula for a parallelogram is 𝐴 = 𝑙 ∙ ℎ. In this parallelogram, the length is the vector 𝐴𝐷 (or you could choose 𝐵𝐶) and the height is 𝐴𝐵 (or you could choose 𝐶𝐷). 10 Therefore the area of the parallelogram will be 𝐴 = 𝐴𝐷 × 𝐴𝐵 . Step 1: Find 𝐴𝐷 and 𝐴𝐵. 𝐴𝐷 = 5 − 5 𝐢 + 0 − 2 𝐣 + 6 − 0 𝐤 = 0𝐢 − 2𝐣 + 6𝐤 𝐴𝐵 = 2 − 5 𝐢 + 6 − 2 𝐣 + 1 − 0 𝐤 = −3𝐢 + 4𝐣 + 1𝐤 Step 2: Find 𝐴𝐷 × 𝐴𝐵. 𝐢
𝐣 𝐤
𝐴𝐷 × 𝐴𝐵 = 0 −2 6 −3 4 1
−2 6
0 6
0 −2
=
𝐢−
𝐣+
𝐤 4 1
−3 1
−3 4
Using our TI Graphing Calculator (see Lesson 3) we have… 11 Don’t forget the minus Therefore, 𝐴𝐷 × 𝐴𝐵 = −26𝐢 − 18𝐣 − 6𝐤 sign for the j component! Step 3: Find 𝐴 = 𝐴𝐷 × 𝐴𝐵 . 𝐴𝐷 × 𝐴𝐵 = (−26)! + (−18)! + (−6)! = 1036 ≈ 32.187 So the area of this parallelogram is approximately 32.187 square units. Finding the Area of Any Figure We can find the area of any shape in the 3D plane. First, simply state your area formula to determine which sides of the figure you will need to find. Second, find the vectors of the corresponding vertices necessary for your sides. Third, find the length (or magnitude) of each of your vectors (sides). Use these values in your area formula. 12