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alexandr402 [8]

3 years ago

12

The surface area, S, of a sphere of radius r feet is S = S(r) = 4πr2. Find the instantaneous rate of change of the surface area

with respect to the radius r at r = 4.
a. 32pi
b. 16pi
c. 64pi
d. 4pi

1 answer:

Ksju [112]3 years ago

8 0

S = 4πr^2
Instantaneous rate of change = dS/dr = 8πr
At r = 4, Instantaneous rate of change = 8π(4) = 32π.

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the answer part b) is
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A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.

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$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}$

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$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}$

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choli [55]

Answer: $6\ \text{miles}$

Step-by-step explanation:

Given

A delivery service charge $\$1.25$ for each delivery

and an additional $\$0.75$ for each mile traveled

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The charge is given by

$\Rightarrow 5.75=1.25+0.75x\\\Rightarrow 4.5=0.75x\\\Rightarrow x=6\ \text{miles}$

Therefore, driver traveled 6 miles for delivery

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3 years ago