Answer to Question #124044 in Calculus for Desmond

Question

Suppose a particle P is moving in the plane so that its coordinates are given by P(x, y),

where x = 4 cos 2t, y = 7 sin 2t.

(i) By finding a, b ∈ R such that

x²/a² + y²/b² = 1, show that P is travelling on an elliptical

path.

(ii) Let L(t) be the distance from P to the origin. Obtain an expression for L(t).

(iii) How fast is the distance between P and the origin changing when t = π/8?

(b) A wire of length 100 centimeters is cut into two pieces. One piece is bent to form a square.

The other piece is bent to form an equilateral triangle. Find the dimensions of the two

pieces of wire so that the sum of the areas of the square and the triangle is minimized.

Answer 1

Solution b.

Length of the wire = 100 cm

No. of pieces = 2

Let the length of one piece be $x$ and the other one be 100 - $x$

Let the perimeter of the square piece be $x$ and let the side of the square piece be 'a'

therefore, 4a = $x$ => a= $x/4$

Area = $a2 = x^2/16$

Let the perimeter of the equilateral triangle = $100 - x$

Let the side of the equilateral triangle be 'b'

therefore, 3b= $100 - x$

or, b = ($100-x) / 3$

Area = 3^0.5 / 4 * $b^2$ =( 3 ^0.5 / 4 )* $(100 - x)^2 / 9$

= (3^0.5 / 36 ) * ($100 - x)^2$

Sum of the areas (A) = ($x^2/16) +$ (3^0.5 / 36) * ($100 - x)^2$

For calculating minima, differentiate A w.r.t x and equate it to 0:

$dA/dx = (2x/16) + ($ 3^0.5 / 36) * 2($100 - x) * (-1)$

Now, $dA/dx = 0$

(2$x /16) - (2 *$ 3^0.5 /36)*(100- $x)$ = 0

or, ($x/8) -$ (3^0.5 /18) * ($100 - x) = 0$

or, $(x/8) +$ (3^0.5 /18)$x=$ (3 ^0.5 * 100)/18

or, $x * 0.2212 = 9.6225$

or, $x = 43.5 cm$

therefore, side of the square = $(x/4) = 10.875 cm$ (Answer)

side of the equilateral triangle = ($100 - x)/3 = 18.833 cm$ (Answer)

Solution a. We know that the parametric equation of an ellipse is $x= a cos \theta$ and $y = b sin\theta$

where $a =$ semi major axis and $b=$ semi minor axis

therefore, comparing with $x= 4 cost 2t$ and $y = 7 sint 2t$ we get $a = 4, b=7 and \theta$ = $2t$

therefore, using L.H.S from equation of an ellipse : ($x^2/a^2) + (y^2/b^2)$

= (16 $cos^22t)/16 +(49sint^22t/49)$

= 1

=R.H.S

Hence, the particle is moving in an elliptical path.

ii) L(t) =( ($x- 0)^2 + (y - 0)^2$ )^0.5

= $((16 cos^22t + 49 sin^2t))$^0.5

=$(16cost^22t + 49 (1- cost ^22t))$ ^0.5

=$(49 - 33 cost^22t)$ ^0.5 (Answer)

iii) Rate of change of L(t) = $dL(t)/dt$

=$d[(49 - 33(1+cost4t)/2)$ ]^0.5 / $dt$

=$d$ $[(98 - 33- 33cos 4t)/2]$ ^0.5 / $dt$

= $d[ (65-33cos 4t)/2]$ ^0.5 /$dt$

=(1/2^0.5) * $d[(65 - 33cos4t)]$ ^0.5/$dt$

=$(1/2$ ^0.5 * $2 * (65 - 33cos4t)$ ^0.5) * $[33*4*sin4t]$

=$66 sin4t/(130-66cos4t)$ ^0.5

Putting $t= \pi$ $/ 8$ or $(\pi$ $/8)*(180/\pi$ $)$ = 22.5 degree

we get $d L(t)/dt = (66 sin(4*22.5))/(130 - 66 cos (4*22.5))$ ^0.5

= 66 / 130^0.5

= 5.788 units/ rad (Answer)