Answer to Question #124031 in Calculus for Nestor Freeman

Question

(a) Suppose a particle P is moving in the plane so that its coordinates are given by P(x,y), where x = 4cos2t, y = 7sin2t. (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

a)

x = 4cos 2t and y = 7sin 2t

=> cos 2t = $\frac{x}{4}$ and sin 2t = $\frac {y}{7}$

We know b cos²2t + sin²2t = 1

So $\frac{x^2}{16}$ + $\frac{y^2}{49}$ = 1 , which is an equation of ellipse.

So P is traveling on a elliptical path.

(ii)

L(t) = Distance of P from origin

=> L(t) = $\sqrt{(4cos 2t - 0)^2 + (7sin 2t - 0)^2}$

=> L(t) = $\sqrt{16cos^2 2t + 49sin^2 2t}$

=> L(t) = $\sqrt{16-16sin^2 2t + 49sin^2 2t}$

=> L(t) = $\sqrt{16 + 33sin^2 2t}$

This is the expression for L(t)

(iii)

L(t) = $\sqrt{16 + 33sin^2 2t}$

Differentiating with respect to t

$\frac{d(L(t))}{dt}$ = $\frac{ 66sin 2tcos2t*2}{2\sqrt{16 + 33sin^2 2t}}$ using $\frac{d(√x)}{dt} = \frac{1}{2√x}$

$\frac{d(L(t))}{dt}$ = $\frac{ 33sin 4t}{\sqrt{16 + 33sin^2 2t}}$

So $[\frac{d(L(t))}{dt}]_{t=π/8}$ = $\frac{ 33sin 4(\frac{π}{8})}{\sqrt{16 + 33sin^2 2(\frac{π}{8}})}$

$=> [\frac{d(L(t))}{dt}]_{t=π/8}$ =$\frac{ 33sin \frac{π}{2}}{\sqrt{16 + 33sin^2 \frac{π}{4}}}$

$=> [\frac{d(L(t))}{dt}]_{t=π/8}$ = $\frac{ 33}{\sqrt{16 + \frac{33}{2}}}$

$=> [\frac{d(L(t))}{dt}]_{t=π/8}$ = $\frac{ 33\sqrt2}{\sqrt{65}}$

$=> [\frac{d(L(t))}{dt}]_{t=π/8}$ = $\frac{ 33\sqrt{130}}{65}$

So L(t) is changing at the rate of $\frac{ 33\sqrt{130}}{65}$ = 5.79 (approximately)

(b)

Let the dimensions of two pieces are x cm(to form an equilateral triangle) and y cm ( to form a square).

So x + y = 100

=> y = 100-x

Side of equilateral triangle is $\frac{x}{3}$ cm and that of square is $\frac{100-x}{4}$ cm

Area of equilateral triangle is $\frac{√3}{4} • \frac{x^2}{9}$ cm²

Area of square is $[\frac{100-x}{4}]^2$ cm²

Sum of areas ,

A = $\frac{√3}{4} • \frac{x^2}{9}$ + $[\frac{100-x}{4}]^2$

Differentiating with respect to x

$\frac{dA}{dx}$ = $\frac{2√3}{4} • \frac{x}{9}$ - $[\frac{2(100-x)}{16}]$

=> $\frac{dA}{dx}$ = $\frac{√3x}{18}$ - $\frac{100-x}{8}$

And $\frac{d^2A}{dx^2}$ = $\frac{√3}{18}$ + $\frac{1}{8}$ > 0

For extreme values of A, $\frac{dA}{dx}$ = 0

=> $\frac{√3x}{18}$ - $\frac{100-x}{8}$ = 0

=> $\frac{4√3x-900+9x}{72}$ = 0

=> (4√3+9)x = 900

=> x = $\frac{900}{4√3+9}$

Since $\frac{d^2A}{dx^2}$ > 0, A is minimum at x = $\frac{900}{4√3+9}$ and then y = 100 - $\frac{900}{4√3+9}$ = $\frac{400√3}{4√3+9}$ cm

Therefore the sum of areas will be minimum when dimensions of two pieces are $\frac{900}{4√3+9}$ cm to form an equilateral triangle and $\frac{400√3}{4√3+9}$ cm to form the square. Approximately the respective dimensions are 56.50 cm and 43.50 cm