A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at a rate of 0.5 m2/sec at what rate is the radius decreasing when the area of the sheet is 12 m2

Respuesta :

Answer:

dx/dt  =  0,04 m/sec

Step-by-step explanation:

Area of the circle is:

A(c) =π*x²      where   x is a radius of the circle

Applying differentiation in relation to time we get:

dA(c)/dt   =  π*2*x* dx/dt    

In this equation we know:

dA(c)/dt  = 0,5 m²/sec

And are looking for dx/dt then

0,5  = 2*π*x*dx/dt    when the area of the sheet is 12 m²  (1)

When  A(c) = 12 m²      x = ??

A(c)  =  12  =  π*x²      ⇒    12  =  3.14* x²    ⇒  12/3.14  =  x²

x²  = 3,82     ⇒   x  = √3,82    ⇒  x = 1,954 m

Finally plugging ths value in equation (1)

0,5  = 6,28*1,954*dx/dt

dx/dt  =  0,5 /12.28

dx/dt  =  0,04 m/sec

The rate at which the radius is decreasing when the area of the sheet is 12 m² is; dr/dt = 0.041 m/s

We are given;

Area of sheet; A = 12 m²

Rate of change of area; dA/dt = 0.5 m²/s

Now, formula for area of the circular sheet is given as;

A = πr²

Thus; 12 = πr²

r = √(12/π)

r = 1.9554 m

Now, we want to find the rate at which the radius is decreasing and so we differentiate both sides of the area formula with respect to t;

dA/dt = 2πr(dr/dt)

Thus;

0.5 = 2π × 1.9554(dr/dt)

dr/dt = 0.5/(2π × 1.9554)

dr/dt = 0.041 m/s

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