#$&* course Mth 277 qa 09_05Section 9.5
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Given Solution: Solving x = 5 t for t we obtain t = x / 5. Substituting this into the second expression we get y = 3 ( x / 5 ) ^ 2 - 6 which simplifies to give us y = 3/25 x^2 - 6. Note that this is a function which can be graphed in the xy plane, forming a parabola. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. Suppose x = 5 t^2 and y = 3 t^4 - 6. Solve x = 5 t^2 for t, then plug this expression in for t in the second equation. Simplify the result. What is your resulting formula for y in terms of x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t= +or- sqrt(x/5) Since we are putting t to an even exponent, we know that the sign does not matter in this case. y=3(x/5)^4-6 y=3/25*x^2-6 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: From the first equation we obtain t = +- sqrt( x / 5 ). Plugging this into the second equation, and noting that wither t = + sqrt(x/5) or - sqrt(x/5) the value of t^4 is x^2 / 25, we get y = 3 ( sqrt(x/5) )^4 - 6 = 3 x^2 / 25 - 6. This is the same function we obtained in the first question. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q003. Suppose x = 5 e^t and y = 3 e^(2t) - 6. Solve x = 5 e^t for t, then plug this expression in for t in the second equation. Simplify the result. What is your resulting formula for y in terms of x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ln(x/5)=t t=ln(x/5) y=3e^(2ln(x/5))-6 y=3/25x-6 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: x = 5 e^t is solved by taking the natural log of both sides, obtaining first ln(x) = ln(5) + ln(e^t). Since the natural log and exponential functions are inverse function, ln(e^t) is just t so we obtain ln(x) = ln(5) + t so that t = ln(x) - ln(5) = ln(x/5). Substituting this into the second equation we get y = 3 e^(4 ln(x/5) ) - 6 = 3 ( e^(ln(x/5) )^2 - 6 = 3 * (x/5)^2 - 6, or y = 3 x^2 / 25 - 6. Note that this is the same function obtained in each of the first three questions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q004. Let `w = 2 `i - 4 `j + 3 `k, `v = 5 `i + 4 `j - 2 `k. Let `r(t) = `w + t * `v. To orient yourself to these vectors you should: Sketch the `w vector, with its initial point at the origin. Then from the tip, or terminal point, of the `w vector, which is located at the point (2, -4, 3), sketch the `v vector. Now answer the following questions: Let P_0 be the terminal point of the `r vector when t = 0, and let P_2 be the terminal point when t = 2. What is the vector from P_0 to P_2, and what is the unit vector in this direction? What is the unit vector in the direction of `v? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `r(t) = `w + t * `v r(0)=w+0*v=2i-4j+3k r(2)=w+2*v=12i+4j-k Then, we will find the vector and divide by its magnitude vector=<10,8,-4> 10/sqrt(180)i+8/sqrt(180)j-4/sqrt(180)k unit vector of v=(5 `i + 4 `j - 2 `k)/sqrt(45) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `w = 2 `i - 4 `j + 3 `k, `v = 5 `i + 4 `j - 2 `k. r(t) = w + t v = (2 + 5 t) i + (-4 + 4 t) j + (3 - 2 t) k so r(0) = 2 i - 4 j + 3 k and r(2) = 12 i + 4 j - k. From P_0 to P_2 the vector is r(2) - r(0) = 10 i + 8 j - 4 k. The unit vector in this direction is (r(2) - r(0)) / || r(2) - r(0) || = (10 i + 8 j - 4 k) / sqrt(180) = .77 i + .6 j - .3 k, very approximately. The unit vector in the direction of v is v / || v|| = (5 `i + 4 `j - 2 `k) / sqrt(45) . This is identical to our previous result (10 i + 8 j - 4 k) / sqrt(180) (to see this just factor sqrt(4) out of the denominator and simplify). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q005. How does the solution to the preceding problem support the contention that all of the points 'traced out' by the tip of the `r(t) vector lie along a single straight line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It shows that it is on the same line because the unit vectors were identical to eachother. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Thus the vector from the t = 0 point to the t = 2 point is parallel to the v vector. There's nothing special about the t = 2 point, so we conjecture a similar result would apply to the t = 0 point and any other point on the graph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn't exactly know how to explain this until I looked at the solution. The solution made sense and considerably helped with my understanding. ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q006. What are the x, y and z coordinates of the tip of the `r(t) vector? Write in the form x = ..., y = ..., z = ... where you fill in the expressions for ... . Solve the resulting equation for t in terms of x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As stated earlier, w = 2 `i - 4 `j + 3 `k, `v = 5 `i + 4 `j - 2 `k, r(t) = `w + t * `v x=5t+2 y=4t-4 z=-2t+3 t=(x-2)/5 t=(y+4)/4 t=-(z-3)/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: r(t) = w + t v = (2 + 5 t) i + (-4 + 4 t) j + (3 - 2 t) k The x, y and z coordinates of the tip of the vector will therefore be x = 2 + 5 t y = -4 + 4 t z = 3 - 2 t Solving the equations for t we get t = (x - 2) / 5 t = (y + 4) / 4 t = -(z - 3) / 2 so we could write (x - 2) / 5 = (y + 4) / 4 = - (z - 3) / 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q009. Suppose (x - 4) / 3 = (y + 2) / 6 = (z - 2) / 9. What values of x, y and z make these three expressions all equal to zero? If it is known that x = 7, what values of y and z make the other two equations true? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since an expression in parenthesis is divided by a non zero number, all we need to do is solve for zero inside the parentheses. x-4=0 x=4 y+2=0 y=-2 z-2=0 z=2 Since the expressions equal and we know the value of x, we can easily find y and z. (7-4)/3=1 (y+2)/6=1 y=4 (z-2)/9=1 z=11 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (x - 4) / 3 = 0 if x = 4 (y + 2) / 6 = if y = -2 (z - 2) / 9 = 0 if z = 2. So the point (4, -2, 2) satisfies these equations. If x = 7 then (x - 4) / 3 = (7 - 4) / 3 = 1. It would then follow that (y + 2) / 6 and (z - 2) / 9 , both being equal to (x - 4) / 3, will also be equal to 1. So we have (y+2) / 6 = 1, with solution y = 4 (z - 2) / 9 = 1, with solution z = 11. Thus the point (7, 4, 11) is also on the line. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q010. Your solution to the preceding gives you the coordinates of two points. If `w is the vector from the origin to the first of these points, and `v the vector from the first point to the second, express `w and `v in terms of their `i, `j and `k components. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Vector w would only be the first points since you subtract from the initial point which is the origin. w=(4,-2,2) 4i-2j+2k The second vector is a little bit more difficult. We need to subtract final point by the initial. (7,4,11)-(4,-2,2)=3i+6j+9k confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: From the origin to the first point (4, -2, 2) the vectors is w = 4 i - 2 j + 2 k. From the first point to the second (7, 4, 11) the vector is v = (7 - 4) i + (4 - (-2) ) j + (11 - 2) k = 3 i + 6 j + 9 k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q011. If `w = x0 `i + y0 `j + z0 `k and `v = a `i + b `j + c `k, then what, in terms of x0, y0, z0, t, a, b and c, is the expression for the vector `r(t) = `w + t `v? What are the expressions for the x, y and z components of this vector? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r(t)=(x0+a*t)i+(y0+b*t)j+(z0+c*t)k The components can easily be seen in this vector. x=x0+a*t y=y0+b*t z=z0+c*t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `r(t) = w + t v = (x0 + a t) i + (y0 + b t) j + (z0 + c t) k so that x = x0 + a t y = y0 + b t z = z0 + c t. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q012. What do you get when you solve the three expressions each for t? What do you get when you set the three resulting expressions equal? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t=(x-x0)/a t=(y-y0)/b t=(z-z0)/c Setting these equal we get: (x-x0)/a=(y-y0)/b=(z-z0)/c confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: From x = x0 + a t y = y0 + b t z = z0 + c t. we get t = (x - x0) / a t = (y - y0) / b t = (z - z0) / c All three expressions are equal to t so they are all equal to one another. Thus (x - x0) / a = (y - y0) / b = (z - z0) / c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q013. Identify the `w and `v vectors for which the equations (x + 3) / 4 = (y - 2) / 5 = (z + 1) / 3 represent the line `w + s `v. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From earlier, we have: t = (x - x0) / a t = (y - y0) / b t = (z - z0) / c Therefore, we can find out some information rather quickly x0=3 y0=-2 z0=1 a=4 b=5 c=3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: As we have seen the general vector r(t) = w + t v has x, y and z coordinates satisfying (x - x0) / a = (y - y0) / b = (z - z0) / c. These equations do not depend on the choice of parameter. The parameter in the expression r(t) = w + t v is t, but the parameter could as well have been s. Had our original expession been r(s) = w + s v we would still have obtained the equations (x - x0) / a = (y - y0) / b = (z - z0) / c. If the equations are (x + 3) / 4 = (y - 2) / 5 = (z + 1) / 3 then we can identify x0 = 3, y0 = -2, z0 = 1, a = 4, b = 5 and c = 3. The meaning is that our equations describe a line through (3, -2, 1) in the direction of the vector 4 i - 2 j + k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q014. Suppose one straight line is represented by parametric equations x(t) = 3 - 5 t, y(t) = 2 + 4 t and z(t) = -6 - 7 t, while another is represented by the equations x(s) = 2 - 3 s, y(s) = 3 + 6 s and z(s) = -3 - 2 s. If the two lines intersect, it means that for some value of t and some value of x, x(t) = x(s), while for the same values y(t) = y(s) and z(t) = z(s). Show that if t = 1 and s = 2, it isn't so. Express the three given conditions for equality as three simultaneous equations. Do you expect the three equations to have a simultaneous solution? Why or why not? If t = 1 and s = 2: our first point is (3 - 5 * 1, 2 + 4 * 1, -6 - 7 * 1) = (-2, 6, -13) and our second point is (2 - 3 * 2, 3 + 6 * 2, -3 - 2 * 2) = (-4, 15, -7). These points are clearly different. The conditions x(t) = x(s), y(t) = y(s), z(t) = z(s) are written out as 3 - 5 t = 2 - 3 s 2 + 4 t = 3 + 6 s -6 - 7 t = -3 - 2 s If we have three equations in three unknowns, we expect to be able to find a solution. If we have three equations in only two unknowns, a solution is not assured. If the numbers in the equations are selected at random, there is a very small chance that there is a solution. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t=1 (3-5,2+4,-6-7)=(-2,6,-13) s=2 (2-6,3+12,-3-4)=(-4,15,-7) It is obvious that this is not the intersection point because the points are not identical. Each component would have to equal for it to be an intersection point. 3-5t=2-3s 2+4t=3+6s -6-7t=-3-2s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q015. If the three simultaneous equations in the preceding problem have a solution, find it. If they don't, prove that they don't. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do not believe there is a solution. I solved for a variable on paper and plugged it into another expression and the numbers did not come out properly.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!