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choli [55]
2 years ago
13

The area of a circular wave expands across a still pond such that its radius increases by 14 cm each second. Write a formula for

the area A of the circle as a function of time t since the wave begins: A=
Mathematics
1 answer:
lara31 [8.8K]2 years ago
4 0

Answer:

A = π(14t)²

Step-by-step explanation:

The radius is increasing at the rate of 14 cm per second.

We need to find the formula for the area A of the circle as the function of time t.

Initial area of the circle,

A = πr², where r is the radius of the circle

Area as a function of t will be :

A = π(14t)²

Here, 14t is the radius of the wave.

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\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\end{array}

\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}
\end{array}

now... your expression is not in vertex form, that's ok... to do a horizontal shift to the left by 4 units, we can simply, add the C  and B component to the "x" variable, C=4, B =1, that way the horizontal shift of C/B or 4/1 is just +4, giving us a horizontal shift to the left

\bf f(x)=x^2-3x-2\impliedby \textit{let's change that for }f(1x+4)
\\\\\\
f(1x+4)=(1x+4)^2-3(1x+4)-2
\\\\\\
f(x+4)=x^2+8x+16-3x-12-2
\\\\\\
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