ENGR 1405 Engineering Mathematics 1

Faculty of Engineering and Applied Science
2000 Fall

Problem Set 9
Solutions

Note: if you see symbols like ³ or Þ in various places, then your browser is not reading the style sheet for this Web page properly (or the Symbol font is not installed on your computer). The translation is:

£ = £ (<=) ³ = ³ (>=)

- = - (minus) Þ = Þ (implies)

p = p (pi) q = q (theta)

k = k (kappa) Ö = Ö (square root)

For the curve defined parametrically by , determine
1. the tangent vector , and
  
  But 1 / Ö t is real only for t > 0
2. the element of arc length "ds".
  
  $ds = sqrt{(4t^3 + (1+t)^4) / (4t^3*(1+t)^4)} dt$
Also:
1. Eliminate the parameter t to find a Cartesian equation of this curve.
  
  y = -1 / Öt
  Þ t = 1 / y²
  Þ x = (1 / y²) / (1 + 1 / y²) = 1 / (y² + 1)
  But y = -1 / Öt < 0. Therefore
  
  x = 1 / (1 + y²); y < 0.

For the curve defined parametrically by , determine
1. the tangent vector , and
2. the element of arc length "ds".
  
  $ds = (2/k^2)*sqrt{k^6 + 1} dk$

For the curve defined parametrically by , determine
1. the tangent vector , and
2. the element of arc length "ds".
  
  $ds = sqrt{tan^2 t + cot^2 t} dt$

For the curve whose Cartesian equation is

x^2/3 + y^2/3 = 1
1. Write down a simple parametric form for this curve (using trigonometric functions) and find
  
  Using the identity cos²q + sin²q = 1, try
  x^2/3 = cos²t Þ x = cos³t and
  y^2/3 = sin²t Þ y = sin³t
  
  (x, y) = (cos³t, sin³t)
  
  Other parameterizations are also possible.
  The entire curve is traversed over any range of parameter values t spanning 2p.
2. the tangent vector , and
  
  T = < -3 cos²t sin t, 3 sin²t cos t > Þ
  
  T = 3 sin t cos t <-cos t, sin t >
3. the element of arc length "ds".
  
  ds/dt = | 3 sin t cos t | Ö ((- cos t)² + (sin t)²) Therefore
  
  ds = 3 | sin t cos t | dt

For the curve of intersection of the plane and the elliptic cylinder defined parametrically by

find:
1. the tangent vector , and
2. the element of arc length "ds".
  
  $ds = 2 sqrt{2(10 sin^2 t + 12 sin t cos t + 5 cos^2 t)} dt$

For the curve defined parametrically by , determine
1. the tangent vector ,
2. the element of arc length "ds" , and
3. the length of the curve from
  
  Throughout the range of integration, t is non-negative, so that | t | = t.
  
  Let u = t² + 4 then du = 2t dt and
  
  L = 56

For the curve defined parametrically by , determine
1. the tangent vector ,
2. the element of arc length "ds", and
3. the length of the curve from
  
  Everywhere in the range of integration sin(2q) is non-negative, so that
  | sin(2q) | = sin(2q).
Also:
1. Eliminate the parameter q to find a Cartesian equation of this curve.
  [Let q be any real number.]
  
  x = cos²q and y = sin²q.
  But we have the identity cos²q + sin²q = 1. Therefore
  x + y = 1
  However, 0 £ cos²q £ 1 and 0 £ sin²q £ 1
  Therefore the full Cartesian equation of this curve is
  
  x + y = 1 for 0 £ x £ 1
2. Hence sketch this curve.
  
  The curve is the segment of the straight line x + y = 1 (slope = -1, axis intercepts = +1) that falls in the first quadrant, including both axis intercepts.
  [An image of the graph is not available yet.]

For the curve defined parametrically by , determine
1. the tangent vector ,
2. the element of arc length "ds" , and
  
  $ds = e^p sqrt{2} dp$
3. the length of the curve from
4. What type of curve is r(p)?
  
  x² + y² = e^2p and
  y / x = tan q
  The curve is therefore a spiral, opening exponentially fast.

Determine the unit tangent vector, , the unit principal normal vector, , and the curvature k for the curve defined parametrically by .

(a)

$T = (1/sqrt{1 + t^2}) < 1, t >$

(b)

$N = (1/sqrt{1 + t^2})< -t, 1 >$

(c)

[Longer] alternative method:

Let and then

and
But

Also:
Eliminate the parameter t to find a Cartesian equation of this curve.
[Let t be any real number.]

Hence sketch this curve.

x = 4t and y = 2t²
Þ t = x/4 and y = 2(x/4)²

The curve is therefore the entire parabola

y = (x²)/8

It opens upward from a vertex at the origin.
Its axis is the y-axis.
It also passes through the points (-1, 1/8) and (1, 1/8).
[An image of the graph is not available yet.]

Determine the unit tangent vector, , the unit principal normal vector, , and the curvature k for the curve defined parametrically by .

(a)

(b)

(c)

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Created 1999 12 07 and modified 2000 12 05 by Dr. G.H. George