HOME WORK

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HOME WORK
(1) The length of a 12-foot by 8-foot rectangle is increasing at a rate of 3 feet per second and the width is decreasing at 2 feet per second (see figure below).

How fast is the perimeter changing?
Solution
Given dl/dt=3
And dw/dt=-2
Then dp when l=12 ft and w=8ft
Perimeter= 2l+2w
Dp/dt=d/dt{2l+2w}
=2dl/dt+2dw/dt=2(3)+2(-2)
=2 ft/sec
How fast is the area changing?
Solution
Given dl/dt=3, dw/dt=-2
Da/dt when l=12 ft , w=8ft
Area=lxwDa/dt=d/dt{lxw}
=Dl/dt*w+dw/dt*l
=3(8)+ 12(-2)
=0 ft2/sec
(c) Find all critical points and local extremes of the following function on the given intervals.
f (x) = 2-x3 on [-2, 1]
Solution
To find the critical points, take derivative:
f’(x)=-3×2
f’(X)= 0
0=-3×2
X=0
For absolute minima and maxima;
f(0)=2-(0)3=2
f(-2)=2-(-2)3=10
f(1)=2-(1)3=1
Therefore: abs minima =1 at x=1
Abs maxima=10 at x=-2
(d) Calculate the limits of the following
limx→0x+5×2
Solution
Divide both numerator and denominator by x2
=lim x->0 1/x+5/x2
Since we have 0 in the denominator, the limit=0
(e) evaluate A'(x) at x = 1, 2, and 3.
A(x) = -3x 2t dtSolution
=d/dx(t2)
At x=1; (12)-(-32)
=-5
At x=2; (22)-(-32)
=-2
At x=3;(32)-(-32)
=12

(f) . Let A(x) represent the area bounded by the graph and the horizontal axis and vertical lines at t=0 and t=x for the graph in Fig. 25. Evaluate A(x) for x = 1, 2, 3, 4, and 5.
Solution
At X=1; A=1*1= 1unit2
X=2; A=1+1.

5=2.5 units2
X=3; A=2.5+(1*2)= 4.5 units2
X=4; A=4.5+1.5=6 units2
X=5; A=6+1=7 units2

Work cited
Anton, Howard, Irl Bivens, and Stephen Davis. Calculus. Vol. 2. Hoboken: Wiley, 2002.