1.
Let r, v and s be the radius, the volume and the surface area of a sphere at any instant t respectively.
Volume of sphere=v=34πr3
Then,
v=34πr3dtdv=4πr2.dtdrgivendtdv=15 cm3/sec andr=10cmTherefore,15=4π×(10)2×dtdrdtdr=400π15=80π3
Now, surface area=s=4πr2
Then,
s=4πr2dtds=8πrdtdror,dtds=8×π×10×80π3[as r=10,dtdr=80π3]dtds=3cm2/sec.Thus, the rate change of surface area 3cm2/sec.
2.
Given, f(x)=xe−x .
(a)
put x=0,y(0)=0
Thus, the y-intercept is y=0.
(b)
For horizontal asymptotes,
limx→∞xe−x=limx→∞exx(intermidiateform)By L’Hospital rule,limx→∞ex1=0
Thus, horizontal asymptotes is y=0.
For vertical asymptotes,
When y → ∞.
Therefore, f(x)=xe−x=exx
When ex=0 then y → to ∞. But ex=0,∀x.
Therefore, there is no vertical asymptote.
(c)
f′(x)=e−x(1−x)Now, f’(x)=0 at x=1.Therefore, intervals are(−∞,1)and(1,∞).(i)In(−∞,1),f′(x)<0⟹f risesand in(1,∞)f′(x)>0⟹ f fall .(ii)Since, there is only one extreme point. And, it is global maxima.
(d)
f′′(x)=e−x(x−2)Now, f”(x)=0 at x=2.Therefore, intervals are(−∞,2)and(2,∞).(i)In(−∞,2),f′′(x)<0⟹f is concave downand in(2,∞)f′′(x)>0⟹ f is concave up.(ii)Infection point is x=2,because there f”=0.And, f changes from concave down to concave up.
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