Number 20 **Chapter 4: Differentiation Rules****20.** The length \( L \) of a rectangle is decreasing at the rate of 3 cm/sec while the width \( W \) is increasing at the rate of 2 cm/sec. When \( L = 12 \) and \( W = 2 \), find the rate of change of:- a) the area,- b) the perimeter,- c) the length of a diagonal.**21.** The function \( V \) whose graph is sketched below gives the volume of air \( V(t) \) that a man has blown into a balloon after \( t \) seconds. ( \( V = \frac{4}{3} \pi r^3 \) ) Approximately how rapidly is the radius changing after 6 seconds?**Graph Description:**The graph is a plot of volume \( V \) in cubic inches versus time \( t \) in seconds. The x-axis is labeled in increments of 18 seconds from 0 to 54 seconds, and the y-axis represents volume in cubic inches. The graph shows a curve that appears to increase initially and then levels off.Note: The specific data points and the exact shape of the curve cannot be determined from the description alone.

Name: Number 20 **Chapter 4: Differentiation Rules****20.** The length \( L
Uploaded: 2024-03-05T11:52:21.000Z
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Description: Number 20 **Chapter 4: Differentiation Rules****20.** The length \( L \) of a rectangle is decreasing at the rate of 3 cm/sec while the width \( W \) is increasing at the rate of 2 cm/sec. When \( L = 12 \) and \( W = 2 \), find the rate of change of

Length of the rectangle = L = 12 cmWidth of the rectangle = B = 2 cmRate at which the length of the reactangle is decreasing = dLdt = - 2 cm/secRate at which the breadth of the rectangle is increasing = dBdt = 2 cm/secNote - Decreasing rate is represented with a negative sign and increasing rate is represented by positive sign.a.)Area of the rectangle - AA = L×BFor rate of change of area, differentiating on both sides w.r.t 't', where t is time - dAdt = L dBdt+B dLdt ddx(u×v) = uddx(v)+vddx(u)dAdt = 12 (2) + 2 (-2)dAdt = 24-4dAdt = 20 cm2/secHence the rate of change of area is 20 cm2/sec.b.) Perimeter of rectange - PP = 2(L+B)For rate of change of Perimeter, differentiating on both sides w.r.t 't', where t is time in seconds - dPdt = 2 dLdt+dBdtdPdt = 2 -2+2dPdt = 0 cm/secHence the rate of change of perimeter is 0 cm/sec. That means the perimeter of the rectangle is not changing with time. c.) Diagonal of the rectangle - DD2 = L2 + B2 (using Pythagorus theorem)For rate of change of diagonal, differentiating on both sides with respect to 't' - 2D(2-1)dDdt = 2L(2-1)dLdt + 2B(2-1)dBdt ddxxn = nx(n-1)2DdDdt = 2LdLdt + 2BdBdt(L2+B2)dDdt = LdLdt + BdBdt122+22dDdt = 12 (-2) + 2 (2)148dDdt = -24+4 = -20dDdt = -20148dDdt = -537 cm/secHence, the rate of change of diagonal of rectangle is -537 cm/sec