please explain A particle is moving along the curve y = . As the particle passes

Question

please explain

A particle is moving along the curve y = . As the particle passes through the point (4, 2), its x-coordinate increases at a rate of 3 cm/s. How fast is the distance from the particle to the origin changing at this instant? cm/s If a rectangle has area A, what are the dimensions which yield the smallest perimeter? List them in non-decreasing order (they may depend on A). If a rectangle has perimeter P, what are the dimensions which yield the largest area? List them in non-decreasing order (they may depend on P).

Explanation / Answer

This equation can be best solved in parametric form.

Let the parameters to the curve be:

x = t^2 and y = t

So, Distance from origin is given by sqrt ( x^2 + y^2) = sqrt (t^4 + t^2)

The rate of change of distance =

ds/dt = [4t^3 + 2t ] / [2sqrt(t^4 + t^2)]

At point (4,2) , value of t = 2 (Since y=t=2)

So, ds/dt = [2t^2 + 1] / [sqrt (t^2 + 1) ]

Put t=2,

ds/dt = 9 / sqrt (5) cm /s = 2.236 cm /s

For second problem,

Area = A = l * b = constant

Perimeter = 2(l+b)

now, l = A/b (Since A is a constant)

Perimeter = 2 (b + A/b) which is a function of b alone. We need to maximise the perimeter, hence, dP/db = 0

So, 2 ( 1 - A/b^2) = 0 [obtained by differentiating both sides]

So, solving this we get,

A/b^2 = 1

b = sqrt (A)

So, l = A/b = A / sqrt (A) = Sqrt (A)

Hence smallest perimeter = 2 (l+b) = 2 [ sqrt (A) +sqrt (A) ] = 4 [sqrt (A) ].

Now, for the third part,

P = 2(l+b) is a constant.

A = l * b

b = (P/2) - l

So, A = [(P/2) - l ] * l

A = Pl/2 - l^2

We wish to maximise the area:

dA/dl = 0

So,

P/2 - 2l = 0

So, P = 4l

Or, 2l +2b= 4l

Or, 2b = 2l ;

b = l.

Which shows that for a fixed perimeter of a rectangle, maximum area is obtained when rectangle is a square i.e length = breadth.

Also, minimum perimeter for a fixed area is for a square.

Hope this helps. Please Rate it ASAP. Thanks.

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