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Luda [366]
3 years ago
7

A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula

Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
5 0

Answer: A. A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}

              B. A'(5) = 1.76 cm/s

Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

A=\pi.r^{2}

So to find the rate of the area:

\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}

\frac{dA}{dr} =2.\pi.r

Using r(t)=3-\frac{363}{(t+11)^{2}}

\frac{dr}{dt}=\frac{726}{(t+11)^{3}}

Then

\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]

\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}

Multipying and simplifying:

\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}

The rate at which the area is increasing is given by expression A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}.

B. At t = 5, rate is:

A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}

A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}

A'(5)=\frac{2408693760\pi}{4294967296}

A'(5)=1.76268

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

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