Answer to Question #186679 in Calculus for Hetisani Sewela

Question

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).

Answer 1

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=$v=\frac{4}{3}​πr^3$

Then,

$v=\frac{4}{3}​πr^3\newline \frac{dv}{dt}=4​πr^2.\frac{dr}{dt}​\newline \text{given} \frac{dv}{dt}​=15 \text { cm}^3 \text{/sec and} r=10 \text{cm}\newline \text{Therefore}, 15=4​π×(10)^2×\frac{dr}{dt}​\newline \frac{dr}{dt}=\frac{15}{400π}=\frac{3}{80π}\newline$

Now, surface area=$s=4πr^2$

Then,

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

2.

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

(a)

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

Thus, the y-intercept is y=0.

(b)

For horizontal asymptotes,

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

Thus, horizontal asymptotes is y=0.

For vertical asymptotes,

When y $\to$ ∞.

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

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

Therefore, there is no vertical asymptote.

(c)

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

(d)

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