Question #186679

How fast is the surface area of a spherical balloon increasing when the radius is 10 cm and

the volume is increasing at 15 cm3/sec? [7]

2. Let f be the function defined by

f (x) = xe-x:

(a) Determine the y–intercept.

(b) Determine the horizontal and vertical asymptotes.

(c) Use the sign pattern for f' (x) to determine

(i) the interval(s) over which f rises and where it falls;

(ii) the local extrema.

(d) Use the sign pattern for f'' (x) to determine

(i) where the graph of f is concave up and where it is concave down;

(ii) the inflection point(s) (if any).


1
Expert's answer
2021-05-07T09:41:03-0400

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=43πr3v=\frac{4}{3}​πr^3

Then,

v=43πr3dvdt=4πr2.drdtgivendvdt=15 cm3/sec andr=10cmTherefore,15=4π×(10)2×drdtdrdt=15400π=380πv=\frac{4}{3}​πr^3\newline \frac{dv}{dt}=4​πr^2.\frac{dr}{dt}​\newline \text{given} \frac{dv}{dt}​=15 \text { cm}^3 \text{/sec and} r=10 \text{cm}\newline \text{Therefore}, 15=4​π×(10)^2×\frac{dr}{dt}​\newline \frac{dr}{dt}=\frac{15}{400π}=\frac{3}{80π}\newline

Now, surface area=s=4πr2s=4πr^2

Then,

s=4πr2dsdt=8πrdrdtor,dsdt=8×π×10×380π[as r=10,drdt=380π]dsdt=3cm2/sec.Thus, the rate change of surface area 3cm2/sec.s=4πr^2\newline \frac{ds}{dt}​=8πr\frac{dr}{dt}​\newline or, \frac{ds}{dt}​=8×π×10×\frac{3}{80π}​ \hspace{1cm} [as\space r=10, \frac{dr}{dt}​=\frac{3}{80π}​]\newline \frac{ds}{dt}​=3 \text{cm}^2\text{/sec}.\newline \text{Thus, the rate change of surface area 3cm}^2\text{/sec.}


2.

Given, f(x)=xexf(x)=xe^{-x} .


(a)

put x=0,y(0)=0put \space x=0,\newline y(0)=0

Thus, the y-intercept is y=0.


(b)

For horizontal asymptotes,

limxxex=limxxex(intermidiateform)By L’Hospital rule,limx1ex=0lim_{x \to ∞} xe^{-x}=lim_{x \to ∞} \frac{x}{e^{x}}\hspace{1cm} ( intermidiate form)\newline \text{By L'Hospital rule,}\newline lim_{x \to ∞} \frac{1}{e^{x}}=0

Thus, horizontal asymptotes is y=0.

For vertical asymptotes,

When y \to ∞.

Therefore, f(x)=xex=xexf(x)=xe^{-x}=\frac{x}{e^x}

When ex=0e^{x} =0 then y \to to ∞. But ex0,xe^{x} \neq 0, \forall x.

Therefore, there is no vertical asymptote.


(c)

f(x)=ex(1x)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.f'(x)=e^{-x}(1-x)\newline \text{Now, f'(x)=0 at x=1.}\newline \text{Therefore, intervals are} (-∞,1) \text{and} (1,∞).\newline (i)\newline In (-∞,1), f'(x)<0 \implies \text{f rises} \newline \text{and in} (1,∞) f'(x)>0 \implies \text{ f fall }.\newline (ii)\newline \text{Since, there is only one extreme point. And, it is global maxima.}


(d)

f(x)=ex(x2)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.f''(x)=e^{-x}(x-2)\newline \text{Now, f''(x)=0 at x=2.}\newline \text{Therefore, intervals are} (-∞,2) \text{and} (2,∞).\newline (i)\newline In (-∞,2), f''(x)<0 \implies \text{f is concave down} \newline \text{and in} (2,∞) f''(x)>0 \implies \text{ f is concave up}.\newline (ii)\newline \text{Infection point is x=2,because there f''=0.} \newline \text{And, f changes from concave down to concave up.}

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