On a computer screen a rectangle Is changing shape. At a certain Instant the rec
ID: 2881856 • Letter: O
Question
On a computer screen a rectangle Is changing shape. At a certain Instant the rectangle is 10 cm wide and the (horizontal) width, w, is increasing at the rate of 2 cm/min. At that same instant the rectangle's (vertical) height is 15 cm high, and the height, h, is shrinking at the rate of -4 cm/min. Answer each question using the appropriate units. What is the rate of change of the perimeter, P? What is the rate of change of the area. A? What is the rate of change of the length, L, of a diagonal? What is the rate of change of the slope, m, of a diagonal? (Use the diagonal from lower left to upper right, so m > 0.) What is the rate of change of the angle theta between the diagonal and the horizontal side? What is the rate of change of cos theta? What is the rate of change of ln(area/perimeter)? (natural logarithm) perimeter) What is the rate of change of ln(area)? (natural logarithm)Explanation / Answer
at any time t , h = 15 cm and w = 10 cm
dw/dt = 2 cm/min and dh/dt = - 4 cm/min
a> Perimeter = P(w,h)= 2(w+h)
rate of change of perimeter = dP/dt= 2( dw/dt + dh/dt)= 2(2 - 4) = -4 cm /min
the perimeter is decreasing at a rate of 4 cm/min
b> Area = A(w,h) = w*h
dA/dt = dw/dt*h + dh/dt*w = 2*15 + (-4)*10 = -10 cm^2/min
=> the area is decreasing at a rate of 10 cm^2/min
c> L (w,h) = sqrt(w^2+h^2)
dL/dt = 1/(2sqrt(w^2+h^2)) * (2w*dw/dt + 2h*dh/dt)
dL/dt = 1/(2sqrt(10^2+15^2)) * (2*10*2 + 2*15*(-4)) = - 8*sqrt(13)/13 cm/min
the diagonal length is decreasing at a rate of 8*sqrt(13)/13 cm/min
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