Section 9.5

Question:  `q001.  Suppose x = 5 t and y = 3 t^2 - 6. 

Solve x = 5 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?

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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):

 

 

 

 

 

 

 

 

 

 

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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?

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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.

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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?

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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):

 

 

 

 

 

 

 

 

 

 

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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?

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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).

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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?

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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.

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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.

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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.

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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?

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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):

 

 

 

 

 

 

 

 

 

 

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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.

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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.

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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? 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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.

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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?

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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.

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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.

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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):

 

 

 

 

 

 

 

 

 

 

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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.

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Given Solution: 

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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.

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Given Solution: 

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