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On a computer screen a rectangle is changing shape. At a certain instant the rec

ID: 2878515 • 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. Some earlier rates can be used in subsequent questions. 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? (m = tan theta = h/w) What is the rate of change of the angle theta between the diagonal and the horizontal side? What is the rate of change of cost het? What is the rate of change of ln (area/perimeter)? (natural logarithm) What is the rate of change of ln(area)? (natural logarithm) For the function f(x) = x^2 - 2x + 5 over [0, 2], calculate R_4 (Approximating area using 4 rectangles using the right end points), calculate the Riemann sum R_n using the right end points, compute the area under the curve f (x) = x^2 - 2x + 5 over [0, 2] given by A = lim_n rightarrow infinity R_n, where R_n = Delta x sigma_i = 1^n f (x_i), Delta x = b - a/n, X_i = X_0 + i Delta x.

Explanation / Answer

1)

given

w =10cm, dw/dt =2cm/min

h=15cm , dh/dt =-4cm/min(negative because it is shrinking)

a)perimeter P =2(w+h)

dP/dt =2(dw/dt +dh/dt)

dP/dt =2(2-4)

dP/dt =-4

perimeter shrinking at 4 cm/min

b)area A=wh

dA/dt =(dw/dt)h +w(dh/dt)

dA/dt=(2*15)+(10*-4)

dA/dt=30-40

dA/dt=-10

area shrinking at 10 cm2/min

c)

length of diagonal L =(w2+h2)

dL/dt =(1/2(w2+h2))(2w(dw/dt) +2h(dh/dt))

dL/dt =((w(dw/dt) +h(dh/dt))/(w2+h2))

dL/dt =(((10*2) +(15*-4))/(102+152))

dL/dt =(-40/325)

dL/dt =(-8/13)

diagonal shrinking at 8/13 cm/min

d)m =h/w

dm/dt =[(dh/dt)w -h(dw/dt)]/w2

dm/dt =[(-4*10) -(15*2)]/102

dm/dt =-30/100

dm/dt =-3/10

slope decreasing at 3/10 units per minute

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