Approximate Value Questions in English

(B) We can write $(7.995)^{1/3}$ as $(8 - 0.005)^{1/3}$.
Using the binomial expansion $(1 - x)^n \approx 1 - nx + \frac{n(n-1)}{2!}x^2$ for small $x$:
$(8 - 0.005)^{1/3} = 8^{1/3} \left(1 - \frac{0.005}{8}\right)^{1/3} = 2 \left(1 - \frac{0.005}{8}\right)^{1/3}$.
Applying the expansion $(1 - x)^n \approx 1 - nx$:
$2 \left(1 - \frac{1}{3} \times \frac{0.005}{8}\right) = 2 \left(1 - \frac{0.005}{24}\right)$.
$2 \left(1 - 0.00020833...\right) = 2(0.99979166...)$.
$= 1.9995833...$.
Rounding to four decimal places,we get $1.9996$.

(A) The area of a circle is given by $A = \pi r^2$.
Given that the radius $r$ increases from $r_1 = 3 \, cm$ to $r_2 = 3.2 \, cm$.
The change in radius is $\Delta r = 3.2 - 3 = 0.2 \, cm$.
The increase in the area of the circle is $\Delta A = A(r_2) - A(r_1) = \pi(r_2^2 - r_1^2)$.
Substituting the values,we get $\Delta A = \pi((3.2)^2 - (3)^2)$.
$\Delta A = \pi(10.24 - 9) = 1.24\pi \, cm^2$.
Therefore,the correct option is $A$.

(B) Given the function $y = x^3 + 5$.
We need to find the approximate change in $y$ $(dy)$ when $x$ changes from $x = 3$ to $x = 2.99$.
Here,$x = 3$ and the change in $x$ is $dx = 2.99 - 3 = -0.01$.
The derivative of $y$ with respect to $x$ is $\frac{dy}{dx} = 3x^2$.
Therefore,the approximate change $dy$ is given by $dy = (3x^2) \cdot dx$.
Substituting the values $x = 3$ and $dx = -0.01$:
$dy = 3(3)^2 \cdot (-0.01)$
$dy = 3(9) \cdot (-0.01)$
$dy = 27 \cdot (-0.01)$
$dy = -0.27$.
Thus,the approximate change in $y$ is $-0.27$.

(A) Let $y = x^{1/3}$,where $x = 125$ and $x + \Delta x = 127$.
Then $\Delta x = 2$.
The derivative is $\frac{dy}{dx} = \frac{1}{3x^{2/3}}$.
For $x = 125$,$y = (125)^{1/3} = 5$ and $\frac{dy}{dx} = \frac{1}{3(125)^{2/3}} = \frac{1}{3(25)} = \frac{1}{75}$.
Using the approximation $\Delta y \approx \frac{dy}{dx} \cdot \Delta x$,we get $\Delta y \approx \frac{1}{75} \times 2 = \frac{2}{75} = 0.02666... \approx 0.0267$.
Therefore,$(127)^{1/3} = y + \Delta y = 5 + 0.0267 = 5.0267$.

(A) Let $y = f(x) = \sqrt{x}$.
We choose $x = 36$ and $\Delta x = 0.6$ such that $x + \Delta x = 36.6$.
We know that $dy \approx \Delta y$,where $dy = f'(x) \Delta x$.
First,find the derivative: $f'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$.
At $x = 36$,$f'(36) = \frac{1}{2\sqrt{36}} = \frac{1}{2 \times 6} = \frac{1}{12}$.
Now,calculate the differential $dy$:
$dy = f'(36) \Delta x = \frac{1}{12} \times 0.6 = \frac{0.6}{12} = 0.05$.
Since $f(x + \Delta x) \approx f(x) + dy$,we have:
$\sqrt{36.6} \approx \sqrt{36} + 0.05 = 6 + 0.05 = 6.05$.

(B) Let $y = x^{\frac{1}{3}}$. Let $x = 27$ and $\Delta x = -2$.
Then $\Delta y = (x + \Delta x)^{\frac{1}{3}} - x^{\frac{1}{3}} = (25)^{\frac{1}{3}} - (27)^{\frac{1}{3}} = (25)^{\frac{1}{3}} - 3$.
So,$(25)^{\frac{1}{3}} = 3 + \Delta y$.
Now,$dy$ is approximately equal to $\Delta y$,given by $dy = \left( \frac{dy}{dx} \right) \Delta x$.
Since $y = x^{\frac{1}{3}}$,$\frac{dy}{dx} = \frac{1}{3x^{\frac{2}{3}}}$.
Substituting $x = 27$ and $\Delta x = -2$:
$dy = \frac{1}{3(27)^{\frac{2}{3}}} \times (-2) = \frac{-2}{3(3^3)^{\frac{2}{3}}} = \frac{-2}{3(3^2)} = \frac{-2}{27} \approx -0.074$.
Thus,the approximate value is $3 + (-0.074) = 2.926$.

(A) Let $x=3$ and $\Delta x=0.02$. Then,the approximate value is given by $f(x+\Delta x) \approx f(x) + f'(x) \Delta x$.
First,calculate $f(x)$ at $x=3$:
$f(3) = 3(3)^{2} + 5(3) + 3 = 3(9) + 15 + 3 = 27 + 15 + 3 = 45$.
Next,find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(3x^{2} + 5x + 3) = 6x + 5$.
Now,calculate $f'(x)$ at $x=3$:
$f'(3) = 6(3) + 5 = 18 + 5 = 23$.
Using the approximation formula:
$f(3.02) \approx f(3) + f'(3) \Delta x$
$f(3.02) \approx 45 + (23)(0.02)$
$f(3.02) \approx 45 + 0.46 = 45.46$.
Thus,the approximate value of $f(3.02)$ is $45.46$.

(A) The volume of a cube with side $x$ is given by $V = x^{3}$.
The approximate change in volume $dV$ is given by the formula $dV = \left(\frac{dV}{dx}\right) \Delta x$.
First,differentiate $V$ with respect to $x$:
$\frac{dV}{dx} = 3x^{2}$.
Given that the side $x$ is increased by $2 \%$,the change in side $\Delta x = 2 \% \text{ of } x = 0.02x$.
Now,substitute these values into the formula for $dV$:
$dV = (3x^{2}) \times (0.02x) = 0.06x^{3} \, m^{3}$.
Thus,the approximate change in volume is $0.06x^{3} \, m^{3}$.

(A) Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius.
Given $r = 9 \text{ cm}$ and $\Delta r = 0.03 \text{ cm}$.
The volume $V$ of the sphere is given by $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $r$,we get $\frac{dV}{dr} = 4 \pi r^2$.
The approximate error in volume $dV$ is given by $dV = \left( \frac{dV}{dr} \right) \Delta r$.
Substituting the values,$dV = (4 \pi r^2) \Delta r = 4 \pi (9)^2 (0.03)$.
$dV = 4 \pi (81) (0.03) = 324 \pi (0.03) = 9.72 \pi \text{ cm}^3$.
Thus,the approximate error in calculating the volume is $9.72 \pi \text{ cm}^3$.

(B) Let $y = \sqrt{x}$.
We choose $x = 25$ and $\Delta x = 0.3$,so that $x + \Delta x = 25.3$.
Then,$\Delta y = \sqrt{x + \Delta x} - \sqrt{x} = \sqrt{25.3} - \sqrt{25} = \sqrt{25.3} - 5$.
This implies $\sqrt{25.3} = 5 + \Delta y$.
Using the approximation $dy \approx \Delta y$,we have $dy = \left(\frac{dy}{dx}\right) \Delta x$.
Since $y = x^{1/2}$,$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.
Substituting the values,$dy = \frac{1}{2\sqrt{25}} \times 0.3 = \frac{1}{2 \times 5} \times 0.3 = \frac{0.3}{10} = 0.03$.
Therefore,$\sqrt{25.3} \approx 5 + 0.03 = 5.03$.

(B) Consider the function $y = \sqrt{x}$. Let $x = 49$ and $\Delta x = 0.5$.
Then,$\Delta y = \sqrt{x + \Delta x} - \sqrt{x} = \sqrt{49.5} - \sqrt{49} = \sqrt{49.5} - 7$.
This implies $\sqrt{49.5} = 7 + \Delta y$.
We know that $dy$ is approximately equal to $\Delta y$,where $dy = \left(\frac{dy}{dx}\right) \Delta x$.
Since $y = \sqrt{x}$,we have $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.
Substituting the values,$dy = \frac{1}{2\sqrt{49}} \times 0.5 = \frac{1}{2 \times 7} \times 0.5 = \frac{0.5}{14} = \frac{1}{28} \approx 0.0357$.
Therefore,the approximate value of $\sqrt{49.5} = 7 + 0.0357 = 7.0357 \approx 7.036$.

(A) Let $f(x) = \sqrt{x}$. We want to find the approximate value of $f(0.6)$.
Choose $x = 0.64$ and $\Delta x = -0.04$,so that $x + \Delta x = 0.6$.
Using the formula $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$:
$f'(x) = \frac{1}{2\sqrt{x}}$.
At $x = 0.64$,$f(0.64) = \sqrt{0.64} = 0.8$.
$f'(0.64) = \frac{1}{2\sqrt{0.64}} = \frac{1}{2(0.8)} = \frac{1}{1.6} = 0.625$.
Therefore,$f(0.6) \approx 0.8 + (0.625)(-0.04)$.
$f(0.6) \approx 0.8 - 0.025 = 0.775$.
The closest option is $0.774$.

(A) Consider the function $y = x^{\frac{1}{3}}$.
Let $x = 0.008$ and $\Delta x = 0.001$,so that $x + \Delta x = 0.009$.
We know that $(0.008)^{\frac{1}{3}} = 0.2$.
Using the differential formula,$\Delta y \approx dy = \left( \frac{dy}{dx} \right) \Delta x$.
Since $y = x^{\frac{1}{3}}$,we have $\frac{dy}{dx} = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$.
At $x = 0.008$,$\frac{dy}{dx} = \frac{1}{3(0.008)^{\frac{2}{3}}} = \frac{1}{3(0.2)^2} = \frac{1}{3(0.04)} = \frac{1}{0.12} = \frac{100}{12} = \frac{25}{3} \approx 8.333$.
Now,$dy = \left( \frac{25}{3} \right) \times 0.001 = \frac{0.025}{3} \approx 0.00833$.
The approximate value is $y + dy = 0.2 + 0.00833 = 0.20833$.

(A) Let $f(x) = x^{\frac{1}{10}}$.
We need to find the value of $f(0.999)$.
Let $x = 1$ and $\Delta x = -0.001$,so that $x + \Delta x = 0.999$.
Using the formula for approximation,$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.
Here,$f(x) = x^{\frac{1}{10}}$,so $f(1) = 1^{\frac{1}{10}} = 1$.
The derivative is $f'(x) = \frac{1}{10} x^{\frac{1}{10} - 1} = \frac{1}{10} x^{-\frac{9}{10}} = \frac{1}{10 x^{\frac{9}{10}}}$.
At $x = 1$,$f'(1) = \frac{1}{10(1)^{\frac{9}{10}}} = \frac{1}{10} = 0.1$.
Now,$f(0.999) \approx f(1) + f'(1) \Delta x$.
$f(0.999) \approx 1 + (0.1)(-0.001) = 1 - 0.0001 = 0.9999$.

(A) Let $y = x^{\frac{1}{4}}$. We want to find the approximate value of $(15)^{\frac{1}{4}}$.
Choose $x = 16$ and $\Delta x = -1$,since $16$ is the nearest fourth power to $15$.
Then $y = (16)^{\frac{1}{4}} = 2$.
The differential $dy$ is given by $dy = \frac{dy}{dx} \Delta x$.
Since $y = x^{\frac{1}{4}}$,$\frac{dy}{dx} = \frac{1}{4} x^{-\frac{3}{4}} = \frac{1}{4 x^{\frac{3}{4}}}$.
At $x = 16$,$\frac{dy}{dx} = \frac{1}{4 (16)^{\frac{3}{4}}} = \frac{1}{4 \times 8} = \frac{1}{32} = 0.03125$.
Thus,$dy = \frac{dy}{dx} \Delta x = 0.03125 \times (-1) = -0.03125$.
The approximate value is $y + dy = 2 + (-0.03125) = 1.96875$.

(B) Let $f(x) = x^{\frac{1}{3}}$.
We know that $27^{\frac{1}{3}} = 3$.
Let $x = 27$ and $\Delta x = -1$.
Then $f(x + \Delta x) = (x + \Delta x)^{\frac{1}{3}} = (26)^{\frac{1}{3}}$.
The approximate value is given by $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.
Here,$f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3(x^{\frac{1}{3}})^2}$.
For $x = 27$,$f'(27) = \frac{1}{3(27^{\frac{1}{3}})^2} = \frac{1}{3(3)^2} = \frac{1}{27}$.
Thus,$(26)^{\frac{1}{3}} \approx 3 + (\frac{1}{27})(-1) = 3 - 0.037037... \approx 2.96296...$
Rounding to four decimal places,we get $2.9630$.

(A) Let $f(x) = x^{\frac{1}{4}}$.
We know that $256^{\frac{1}{4}} = 4$.
Let $x = 256$ and $\Delta x = -1$.
Then,$f(x + \Delta x) = (x + \Delta x)^{\frac{1}{4}} = (255)^{\frac{1}{4}}$.
The approximate value is given by $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.
Here,$f'(x) = \frac{1}{4} x^{-\frac{3}{4}} = \frac{1}{4 x^{\frac{3}{4}}}$.
For $x = 256$,$f'(256) = \frac{1}{4 (256)^{\frac{3}{4}}} = \frac{1}{4 (4^4)^{\frac{3}{4}}} = \frac{1}{4 \times 4^3} = \frac{1}{4 \times 64} = \frac{1}{256} = 0.00390625$.
Thus,$f(255) \approx f(256) + f'(256) \Delta x = 4 + (0.00390625)(-1) = 4 - 0.00390625 = 3.99609375$.
Rounding to four decimal places,we get $3.9961$.

(A) Let $y = x^{\frac{1}{4}}$.
Consider $x = 81$ and $\Delta x = 1$.
Then $y = (81)^{\frac{1}{4}} = 3$.
We know that $\Delta y \approx dy = \left(\frac{dy}{dx}\right) \Delta x$.
Since $y = x^{\frac{1}{4}}$,we have $\frac{dy}{dx} = \frac{1}{4} x^{-\frac{3}{4}} = \frac{1}{4 x^{\frac{3}{4}}}$.
Substituting the values,$dy = \frac{1}{4(81)^{\frac{3}{4}}} \times 1 = \frac{1}{4(3^4)^{\frac{3}{4}}} = \frac{1}{4(3^3)} = \frac{1}{4 \times 27} = \frac{1}{108}$.
Calculating the value,$dy \approx 0.00926$.
Thus,$(82)^{\frac{1}{4}} = y + \Delta y \approx 3 + 0.00926 = 3.00926$.
Rounding to three decimal places,the approximate value is $3.009$.

(A) Let $y = x^{\frac{1}{2}}$.
We choose $x = 400$ and $\Delta x = 1$ such that $x + \Delta x = 401$.
Then,$\Delta y = \sqrt{x + \Delta x} - \sqrt{x} = \sqrt{401} - \sqrt{400} = \sqrt{401} - 20$.
This implies $\sqrt{401} = 20 + \Delta y$.
Using the approximation $dy \approx \Delta y$,we have $dy = \left(\frac{dy}{dx}\right) \Delta x$.
Since $y = x^{\frac{1}{2}}$,we have $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.
Substituting the values,$dy = \frac{1}{2\sqrt{400}} \times 1 = \frac{1}{2 \times 20} = \frac{1}{40} = 0.025$.
Therefore,the approximate value of $\sqrt{401} = 20 + 0.025 = 20.025$.

(A) Let $y = x^{\frac{1}{2}}$.
We choose $x = 0.0036$ and $\Delta x = 0.0001$ such that $x + \Delta x = 0.0037$.
Then $\Delta y = (x + \Delta x)^{\frac{1}{2}} - x^{\frac{1}{2}} = (0.0037)^{\frac{1}{2}} - (0.0036)^{\frac{1}{2}} = (0.0037)^{\frac{1}{2}} - 0.06$.
So,$(0.0037)^{\frac{1}{2}} = 0.06 + \Delta y$.
We know that $dy \approx \Delta y$,where $dy = \left(\frac{dy}{dx}\right) \Delta x$.
Since $y = x^{\frac{1}{2}}$,$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.
Thus,$dy = \frac{1}{2\sqrt{0.0036}} \times 0.0001 = \frac{1}{2 \times 0.06} \times 0.0001 = \frac{0.0001}{0.12} = \frac{1}{1200} \approx 0.000833$.
Therefore,$(0.0037)^{\frac{1}{2}} \approx 0.06 + 0.000833 = 0.060833$.

(A) Let $y = x^{\frac{1}{3}}$.
Consider $x = 27$ and $\Delta x = -0.43$,so that $x + \Delta x = 26.57$.
Then,$\Delta y = (x + \Delta x)^{\frac{1}{3}} - x^{\frac{1}{3}} = (26.57)^{\frac{1}{3}} - (27)^{\frac{1}{3}} = (26.57)^{\frac{1}{3}} - 3$.
This implies $(26.57)^{\frac{1}{3}} = 3 + \Delta y$.
We use the approximation $dy \approx \Delta y$,where $dy = \left(\frac{dy}{dx}\right) \Delta x$.
Since $y = x^{\frac{1}{3}}$,we have $\frac{dy}{dx} = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3(x^{\frac{1}{3}})^2}$.
At $x = 27$,$\frac{dy}{dx} = \frac{1}{3(3)^2} = \frac{1}{27}$.
Thus,$dy = \left(\frac{1}{27}\right) (-0.43) = -\frac{0.43}{27} \approx -0.0159$.
Therefore,$(26.57)^{\frac{1}{3}} \approx 3 - 0.0159 = 2.9841$.
The approximate value is $2.984$.

(A) Let $y = x^{\frac{1}{4}}$. Consider $x = 81$ and $\Delta x = 0.5$.
Then,$\Delta y = (x + \Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} = (81.5)^{\frac{1}{4}} - (81)^{\frac{1}{4}} = (81.5)^{\frac{1}{4}} - 3$.
This implies $(81.5)^{\frac{1}{4}} = 3 + \Delta y$.
We know that $dy \approx \Delta y$,where $dy = \left( \frac{dy}{dx} \right) \Delta x$.
Since $y = x^{\frac{1}{4}}$,we have $\frac{dy}{dx} = \frac{1}{4} x^{-\frac{3}{4}} = \frac{1}{4x^{\frac{3}{4}}}$.
Substituting the values,$dy = \frac{1}{4(81)^{\frac{3}{4}}} \times 0.5 = \frac{0.5}{4 \times (3^4)^{\frac{3}{4}}} = \frac{0.5}{4 \times 3^3} = \frac{0.5}{4 \times 27} = \frac{0.5}{108} \approx 0.0046$.
Therefore,the approximate value is $3 + 0.0046 = 3.0046$.

(A) Let $y = x^{\frac{3}{2}}$.
Consider $x = 4$ and $\Delta x = -0.032$.
Then,$\Delta y = (x + \Delta x)^{\frac{3}{2}} - x^{\frac{3}{2}} = (3.968)^{\frac{3}{2}} - (4)^{\frac{3}{2}} = (3.968)^{\frac{3}{2}} - 8$.
So,$(3.968)^{\frac{3}{2}} = 8 + \Delta y$.
Now,$dy$ is approximately equal to $\Delta y$,given by $dy = \left(\frac{dy}{dx}\right) \Delta x$.
Since $y = x^{\frac{3}{2}}$,we have $\frac{dy}{dx} = \frac{3}{2} x^{\frac{1}{2}}$.
Thus,$dy = \frac{3}{2} (4)^{\frac{1}{2}} (-0.032) = \frac{3}{2} (2) (-0.032) = 3 (-0.032) = -0.096$.
Therefore,the approximate value is $8 + (-0.096) = 7.904$.

(A) Let $y = x^{\frac{1}{5}}$.
Let $x = 32$ and $\Delta x = 0.15$.
Then,$\Delta y = (x + \Delta x)^{\frac{1}{5}} - x^{\frac{1}{5}} = (32.15)^{\frac{1}{5}} - (32)^{\frac{1}{5}} = (32.15)^{\frac{1}{5}} - 2$.
So,$(32.15)^{\frac{1}{5}} = 2 + \Delta y$.
Now,$dy$ is approximately equal to $\Delta y$,given by $dy = \left(\frac{dy}{dx}\right) \Delta x$.
Since $y = x^{\frac{1}{5}}$,we have $\frac{dy}{dx} = \frac{1}{5} x^{-\frac{4}{5}} = \frac{1}{5(x^{\frac{1}{5}})^4}$.
Substituting $x = 32$ and $\Delta x = 0.15$:
$dy = \frac{1}{5(32^{\frac{1}{5}})^4} \cdot (0.15) = \frac{1}{5(2)^4} \cdot (0.15) = \frac{0.15}{5 \cdot 16} = \frac{0.15}{80} = 0.001875$.
Rounding to five decimal places,$dy \approx 0.001875$.
Thus,$(32.15)^{\frac{1}{5}} \approx 2 + 0.001875 = 2.001875$.

(A) Let $x=2$ and $\Delta x=0.01.$ Then,we have:
$f(x+\Delta x) \approx f(x) + f'(x) \Delta x$
Given $f(x) = 4x^{2} + 5x + 2$,we find the derivative:
$f'(x) = 8x + 5$
At $x=2$,$f(2) = 4(2)^{2} + 5(2) + 2 = 16 + 10 + 2 = 28$.
At $x=2$,$f'(2) = 8(2) + 5 = 16 + 5 = 21$.
Now,substitute these values into the approximation formula:
$f(2.01) \approx f(2) + f'(2) \Delta x$
$f(2.01) \approx 28 + (21)(0.01)$
$f(2.01) \approx 28 + 0.21$
$f(2.01) \approx 28.21$
Thus,the approximate value of $f(2.01)$ is $28.21$.

(A) Let $x=5$ and $\Delta x=0.001$. Then,we have:
$f(5.001) = f(x+\Delta x) \approx f(x) + f'(x) \Delta x$.
First,calculate $f(x)$ at $x=5$:
$f(5) = (5)^{3} - 7(5)^{2} + 15 = 125 - 175 + 15 = -35$.
Next,find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(x^{3} - 7x^{2} + 15) = 3x^{2} - 14x$.
Calculate $f'(x)$ at $x=5$:
$f'(5) = 3(5)^{2} - 14(5) = 3(25) - 70 = 75 - 70 = 5$.
Now,use the approximation formula:
$f(5.001) \approx f(5) + f'(5) \Delta x$
$f(5.001) \approx -35 + (5)(0.001)$
$f(5.001) \approx -35 + 0.005 = -34.995$.
Hence,the approximate value of $f(5.001)$ is $-34.995$.

(C) The volume $V$ of a cube with side $x$ is given by $V = x^{3}$.
The approximate change in volume $dV$ is given by the formula $dV = \left(\frac{dV}{dx}\right) \Delta x$.
First,differentiate $V$ with respect to $x$:
$\frac{dV}{dx} = 3x^{2}$.
Given that the side $x$ increases by $1 \%$,the change in side $\Delta x$ is $1 \% \text{ of } x = 0.01x$.
Now,substitute these values into the formula for $dV$:
$dV = (3x^{2}) \times (0.01x)$.
$dV = 0.03x^{3}$.
Therefore,the approximate change in the volume of the cube is $0.03 \,x^{3} \, m^{3}$.

(B) The surface area $S$ of a cube with side $x$ is given by $S = 6x^{2}.$
To find the approximate change in surface area $(\Delta S)$,we use the differential formula $\Delta S \approx \frac{dS}{dx} \cdot \Delta x.$
First,differentiate $S$ with respect to $x$:
$\frac{dS}{dx} = \frac{d}{dx}(6x^{2}) = 12x.$
The side is decreased by $1\%$,so the change in side $\Delta x = -0.01x.$
Now,calculate the approximate change:
$\Delta S \approx (12x) \cdot (-0.01x) = -0.12x^{2}.$
The magnitude of the approximate change in the surface area is $0.12x^{2} \, m^{2}.$

(A) Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius.
Given, $r = 7 \text{ m}$ and $\Delta r = 0.02 \text{ m}$.
The volume $V$ of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $r$, we get $\frac{dV}{dr} = 4 \pi r^2$.
The approximate error in volume $\Delta V$ is given by $\Delta V \approx \frac{dV}{dr} \cdot \Delta r$.
Substituting the values, $\Delta V \approx (4 \pi r^2) \cdot \Delta r$.
$\Delta V \approx 4 \pi (7)^2 \cdot (0.02) = 4 \pi (49) (0.02) = 196 \pi (0.02) = 3.92 \pi \text{ m}^3$.
Thus, the approximate error in calculating the volume is $3.92 \pi \text{ m}^3$.

(A) Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius. Given $r = 9 \text{ m}$ and $\Delta r = 0.03 \text{ m}$.
The surface area $S$ of a sphere is given by $S = 4 \pi r^2$.
Differentiating $S$ with respect to $r$,we get $\frac{dS}{dr} = 8 \pi r$.
The approximate error in the surface area is given by $dS = \left( \frac{dS}{dr} \right) \Delta r$.
Substituting the values,$dS = (8 \pi \times 9) \times 0.03$.
$dS = 72 \pi \times 0.03 = 2.16 \pi \text{ m}^2$.
Thus,the approximate error in calculating the surface area is $2.16 \pi \text{ m}^2$.

(D) Let $x=3$ and $\Delta x=0.02$. Then,we have:
$f(x+\Delta x) \approx f(x) + f'(x) \Delta x$
Given $f(x) = 3x^{2} + 15x + 5$,we find the derivative:
$f'(x) = \frac{d}{dx}(3x^{2} + 15x + 5) = 6x + 15$
Now,substitute $x=3$ and $\Delta x=0.02$ into the approximation formula:
$f(3.02) \approx f(3) + f'(3)(0.02)$
Calculate $f(3)$:
$f(3) = 3(3)^{2} + 15(3) + 5 = 3(9) + 45 + 5 = 27 + 45 + 5 = 77$
Calculate $f'(3)$:
$f'(3) = 6(3) + 15 = 18 + 15 = 33$
Substitute these values back:
$f(3.02) \approx 77 + 33(0.02)$
$f(3.02) \approx 77 + 0.66$
$f(3.02) \approx 77.66$
Thus,the approximate value is $77.66$.

(A) The volume of a cube $(V)$ with side $x$ is given by $V = x^{3}$.
The approximate change in volume $(dV)$ is given by the formula $dV = \left(\frac{dV}{dx}\right) \Delta x$.
First,differentiate $V$ with respect to $x$: $\frac{dV}{dx} = 3x^{2}$.
Given that the side increases by $3 \%$,the change in side $\Delta x = 3 \% \text{ of } x = 0.03x$.
Now,substitute these values into the formula for $dV$:
$dV = (3x^{2}) \times (0.03x)$
$dV = 0.09x^{3} \text{ m}^{3}$.
Thus,the approximate change in the volume of the cube is $0.09x^{3} \text{ m}^{3}$.

(A) Let $y = x^{\frac{1}{4}}$. Let $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$.
Then,$\Delta y = (x + \Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} = \left(\frac{17}{81}\right)^{\frac{1}{4}} - \left(\frac{16}{81}\right)^{\frac{1}{4}} = \left(\frac{17}{81}\right)^{\frac{1}{4}} - \frac{2}{3}$.
Thus,$\left(\frac{17}{81}\right)^{\frac{1}{4}} = \frac{2}{3} + \Delta y$.
Using the differential $dy \approx \Delta y$,we have $dy = \frac{dy}{dx} \Delta x = \frac{1}{4} x^{-\frac{3}{4}} \Delta x$.
Substituting $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$:
$dy = \frac{1}{4 \left(\frac{16}{81}\right)^{\frac{3}{4}}} \times \frac{1}{81} = \frac{1}{4 \times \frac{8}{27}} \times \frac{1}{81} = \frac{27}{32} \times \frac{1}{81} = \frac{1}{32 \times 3} = \frac{1}{96} \approx 0.0104$.
Therefore,the approximate value is $\frac{2}{3} + 0.0104 = 0.6667 + 0.0104 = 0.6771 \approx 0.677$.

(B) Let $y = f(x) = x^{-\frac{1}{5}}$.
Let $x = 32$ and $\Delta x = 1$,so that $x + \Delta x = 33$.
We know that $f(x) = x^{-\frac{1}{5}} = (32)^{-\frac{1}{5}} = (2^5)^{-\frac{1}{5}} = 2^{-1} = \frac{1}{2} = 0.5$.
Using the differential formula,$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.
First,find the derivative: $f'(x) = -\frac{1}{5} x^{-\frac{6}{5}}$.
At $x = 32$,$f'(32) = -\frac{1}{5} (32)^{-\frac{6}{5}} = -\frac{1}{5} (2^5)^{-\frac{6}{5}} = -\frac{1}{5} (2^{-6}) = -\frac{1}{5 \times 64} = -\frac{1}{320}$.
Now,calculate the approximate value: $f(33) \approx 0.5 + (-\frac{1}{320})(1)$.
$f(33) \approx 0.5 - 0.003125 = 0.496875$.
Rounding to three decimal places,we get $0.497$.

(N/A) Let $\triangle ABC$ be a right-angled triangle at $B$. Let $P$ be a point on the hypotenuse $AC$ such that the distance of $P$ from $AB$ is $a$ and from $BC$ is $b$.
Let $\angle C = \theta$. Then $\angle A = 90^{\circ} - \theta$.
In the right-angled triangle formed by the perpendicular from $P$ to $BC$,we have $PC = \frac{b}{\sin \theta} = b \csc \theta$.
In the right-angled triangle formed by the perpendicular from $P$ to $AB$,we have $AP = \frac{a}{\cos \theta} = a \sec \theta$.
The length of the hypotenuse $L = AC = AP + PC = a \sec \theta + b \csc \theta$.
To find the minimum length,we differentiate $L$ with respect to $\theta$:
$\frac{dL}{d\theta} = a \sec \theta \tan \theta - b \csc \theta \cot \theta$.
Setting $\frac{dL}{d\theta} = 0$,we get $a \sec \theta \tan \theta = b \csc \theta \cot \theta$.
$\frac{a \sin \theta}{\cos^2 \theta} = \frac{b \cos \theta}{\sin^2 \theta} \Rightarrow \tan^3 \theta = \frac{b}{a} \Rightarrow \tan \theta = (\frac{b}{a})^{\frac{1}{3}}$.
At this value,$\sin \theta = \frac{b^{\frac{1}{3}}}{(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}}}$ and $\cos \theta = \frac{a^{\frac{1}{3}}}{(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}}}$.
Substituting these into $L = a \sec \theta + b \csc \theta$:
$L = a \frac{(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}}}{a^{\frac{1}{3}}} + b \frac{(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}}}{b^{\frac{1}{3}}} = a^{\frac{2}{3}}(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}} + b^{\frac{2}{3}}(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}}$.
$L = (a^{\frac{2}{3}} + b^{\frac{2}{3}})(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}} = (a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}}$.
Thus,the minimum length of the hypotenuse is $(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}}$.

(N/A) Let $s$ be the curved surface area of the conical water level. The formula for the curved surface area is $s = \pi r l$.
Given the semi-vertical angle $\alpha = \frac{\pi}{4}$,we have $r = l \sin(\alpha) = l \sin(\frac{\pi}{4}) = \frac{l}{\sqrt{2}}$.
Substituting $r$ into the surface area formula: $s = \pi (\frac{l}{\sqrt{2}}) l = \frac{\pi}{\sqrt{2}} l^2$.
Differentiating with respect to time $t$: $\frac{ds}{dt} = \frac{\pi}{\sqrt{2}} \cdot 2l \cdot \frac{dl}{dt} = \sqrt{2} \pi l \frac{dl}{dt}$.
Given $\frac{ds}{dt} = -2 \text{ cm}^2/\text{sec}$ (since the area is decreasing),we have $-2 = \sqrt{2} \pi l \frac{dl}{dt}$.
At $l = 4 \text{ cm}$,$-2 = \sqrt{2} \pi (4) \frac{dl}{dt}$.
Solving for $\frac{dl}{dt}$: $\frac{dl}{dt} = \frac{-2}{4 \sqrt{2} \pi} = -\frac{1}{2 \sqrt{2} \pi} = -\frac{\sqrt{2}}{4 \pi} \text{ cm/sec}$.
Thus,the rate of decrease of the slant height is $\frac{\sqrt{2}}{4 \pi} \text{ cm/sec}$.

(A) We use the formula for linear approximation: $f(x + \Delta x) \approx f(x) + \Delta x \cdot f'(x)$.
Let $f(x) = \sqrt{x}$. Then $f'(x) = \frac{1}{2\sqrt{x}}$.
We choose $x = 0.09$ because $\sqrt{0.09} = 0.3$ is a perfect square root.
Then $\Delta x = 0.082 - 0.09 = -0.008$.
Substituting these values into the formula:
$f(0.082) \approx f(0.09) + (-0.008) \cdot f'(0.09)$
$f(0.082) \approx 0.3 + (-0.008) \cdot \left(\frac{1}{2 \cdot 0.3}\right)$
$f(0.082) \approx 0.3 - \frac{0.008}{0.6}$
$f(0.082) \approx 0.3 - 0.01333...$
$f(0.082) \approx 0.28666... \approx 0.2867$.

(A) Given $y = x^{4} - 10$.
We need to find the approximate change in $y$ $(\Delta y)$ when $x$ changes from $x = 2$ to $x = 1.99$.
Here,$x = 2$ and $\Delta x = 1.99 - 2 = -0.01$.
The derivative is $\frac{dy}{dx} = 4x^{3}$.
The approximate change is given by $\Delta y \approx \frac{dy}{dx} \times \Delta x$.
Substituting the values: $\Delta y \approx 4(2)^{3} \times (-0.01)$.
$\Delta y \approx 4(8) \times (-0.01) = 32 \times (-0.01) = -0.32$.
Thus,the change in $y$ is $-0.32$.

(A) Let $y = f(x) = x^{5}$.
We need to find the value of $(1.999)^{5}$.
Let $x = 2$ and $\Delta x = -0.001$,such that $x + \Delta x = 1.999$.
Then,$f(x) = x^{5} = 2^{5} = 32$.
The derivative is $f'(x) = 5x^{4}$.
Using the formula for approximation,$f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x$.
$f'(2) = 5(2)^{4} = 5 \times 16 = 80$.
Therefore,$f(1.999) \approx 32 + 80 \times (-0.001)$.
$f(1.999) \approx 32 - 0.080 = 31.920$.

(N/A) Let internal radius $= r = 3 \text{ cm}$ and external radius $= R = 3.0005 \text{ cm}$.
The volume of a hollow spherical shell is given by $V = \frac{4}{3} \pi (R^3 - r^3)$.
Let $f(x) = x^3$. Then $R^3 = f(3.0005)$ and $r^3 = f(3) = 3^3 = 27$.
Using the differential approximation,$\Delta V \approx dV = \frac{dV}{dR} \Delta R$.
Here,$V = \frac{4}{3} \pi R^3$,so $\frac{dV}{dR} = 4 \pi R^2$.
Given $R = 3$ and $\Delta R = 0.0005$,we have:
$dV = 4 \pi (3)^2 \times 0.0005$
$dV = 4 \pi \times 9 \times 0.0005$
$dV = 36 \pi \times 0.0005 = 0.018 \pi \text{ cm}^3$.
Thus,the approximate volume of metal is $0.018 \pi \text{ cm}^3$.

(A) Let $f(x) = \cos x$. We need to find the value of $\cos(59^{\circ} 30^{\prime})$.
We can write $59^{\circ} 30^{\prime}$ as $60^{\circ} - 30^{\prime} = 60^{\circ} - 0.5^{\circ}$.
Let $x = 60^{\circ} = \frac{\pi}{3}$ radians and $\Delta x = -0.5^{\circ} = -0.5 \times 0.0175^{c} = -0.00875^{c}$.
Using the differential formula,$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.
Here,$f(x) = \cos x$,so $f'(x) = -\sin x$.
$f(60^{\circ} - 0.5^{\circ}) \approx \cos(60^{\circ}) - \sin(60^{\circ}) \times \Delta x$.
Given $\cos(60^{\circ}) = 0.5$ and $\sin(60^{\circ}) = 0.8660$.
$f(59.5^{\circ}) \approx 0.5 - (0.8660) \times (-0.00875)$.
$f(59.5^{\circ}) \approx 0.5 + 0.0075775 = 0.5075775$.
Rounding to four decimal places,we get $0.5076$.

(C) Let $f(x) = x^{1/3}$. We need to find the approximate value of $f(64.04)$.
Let $x = 64$ and $\Delta x = 0.04$.
Then $f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x$.
Here,$f(x) = x^{1/3}$,so $f'(x) = \frac{1}{3} x^{-2/3} = \frac{1}{3(x^{1/3})^2}$.
For $x = 64$,$f(64) = 64^{1/3} = 4$.
$f'(64) = \frac{1}{3(4)^2} = \frac{1}{3 \cdot 16} = \frac{1}{48}$.
Now,$f(64.04) \approx 4 + \frac{1}{48} \cdot 0.04$.
$f(64.04) \approx 4 + \frac{0.04}{48} = 4 + \frac{4}{4800} = 4 + \frac{1}{1200}$.
$f(64.04) \approx 4 + 0.000833...$
Thus,the approximate value is $4.00083$.

(A) Let $f(x) = \frac{1}{x^2} = x^{-2}$.
We need to find the value of $f(2.002)$.
Let $x = 2$ and $\Delta x = 0.002$.
Then $f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x$.
First,find the derivative $f'(x) = -2x^{-3} = -\frac{2}{x^3}$.
At $x = 2$,$f(2) = \frac{1}{2^2} = \frac{1}{4} = 0.25$.
And $f'(2) = -\frac{2}{2^3} = -\frac{2}{8} = -0.25$.
Now,substitute these values into the approximation formula:
$f(2.002) \approx 0.25 + (-0.25) \cdot (0.002)$.
$f(2.002) \approx 0.25 - 0.0005 = 0.2495$.
Thus,the approximate value is $0.2495$.

(B) Let $f(x) = x^{\frac{1}{3}}$.
Then,$f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$.
We choose $a = 0.027$ because $\sqrt[3]{0.027} = 0.3$,and $h = -0.001$ such that $a + h = 0.026$.
Using the linear approximation formula $f(a + h) \approx f(a) + hf'(a)$:
$f(a) = (0.027)^{\frac{1}{3}} = 0.3$.
$f'(a) = \frac{1}{3(0.027)^{\frac{2}{3}}} = \frac{1}{3(0.09)} = \frac{1}{0.27}$.
Substituting these values:
$f(0.026) \approx 0.3 + (-0.001) \times \frac{1}{0.27}$.
$f(0.026) \approx 0.3 - \frac{0.001}{0.27} \approx 0.3 - 0.0037 = 0.2963$.

(D) Let $f(x) = \cos x$. Then $f^{\prime}(x) = -\sin x$.
We need to find the value of $\cos(30^{\circ} 30^{\prime})$.
Here,$30^{\circ} 30^{\prime} = 30^{\circ} + 30^{\prime} = 30^{\circ} + (0.5)^{\circ}$.
Converting $0.5^{\circ}$ to radians: $0.5 \times 0.0175 = 0.00875 \text{ rad}$.
Let $a = 30^{\circ} = \frac{\pi}{6} \text{ rad}$ and $h = 0.00875 \text{ rad}$.
Using the linear approximation formula $f(a+h) \approx f(a) + h f^{\prime}(a)$:
$f(a+h) \approx \cos(30^{\circ}) + (0.00875)(-\sin(30^{\circ}))$.
Given $\cos 30^{\circ} = 0.8660$ and $\sin 30^{\circ} = 0.5$.
$f(a+h) \approx 0.8660 - 0.00875 \times 0.5$.
$f(a+h) \approx 0.8660 - 0.004375 = 0.861625$.
Rounding to four decimal places,we get $0.8616$.

(B) Let $f(x) = \tan^{-1} x$.
Then $f'(x) = \frac{1}{1+x^2}$.
We use the linear approximation formula $f(a+h) \approx f(a) + h f'(a)$.
Here,let $a = 1$ and $h = -0.001$,so $a+h = 0.999$.
$f(a) = \tan^{-1}(1) = \frac{\pi}{4}$.
$f'(a) = \frac{1}{1+1^2} = \frac{1}{2} = 0.5$.
Substituting these values:
$\tan^{-1}(0.999) \approx \frac{\pi}{4} + (-0.001)(0.5)$.
$\tan^{-1}(0.999) \approx \frac{3.1415}{4} - 0.0005$.
$\tan^{-1}(0.999) \approx 0.785375 - 0.0005$.
$\tan^{-1}(0.999) \approx 0.784875$.
Rounding to four decimal places,we get $0.7849$.

(C) Let $f(x) = x^{3/2}$.
Then,$f'(x) = \frac{3}{2} x^{1/2}$.
We can write $3.978$ as $a + h$,where $a = 4$ and $h = -0.022$.
Using the linear approximation formula $f(a + h) \approx f(a) + h \cdot f'(a)$:
$f(4) = 4^{3/2} = 8$.
$f'(4) = \frac{3}{2} \cdot (4)^{1/2} = \frac{3}{2} \cdot 2 = 3$.
Therefore,$f(3.978) \approx 8 + (-0.022) \cdot 3$.
$f(3.978) \approx 8 - 0.066$.
$f(3.978) \approx 7.934$.

(A) Let $f(x) = x^3-2x^2+3x+2$.
We need to find the approximate value at $x = 2.01$.
Let $a = 2$ and $h = 0.01$,so $x = a+h = 2.01$.
The derivative is $f'(x) = 3x^2-4x+3$.
Calculating $f(a)$ at $a = 2$:
$f(2) = (2)^3-2(2)^2+3(2)+2 = 8-8+6+2 = 8$.
Calculating $f'(a)$ at $a = 2$:
$f'(2) = 3(2)^2-4(2)+3 = 12-8+3 = 7$.
Using the linear approximation formula $f(a+h) \approx f(a) + h \cdot f'(a)$:
$f(2.01) \approx 8 + (0.01)(7) = 8 + 0.07 = 8.07$.

(A) Let $f(x) = x^{\frac{3}{2}}$.
Then,$f'(x) = \frac{3}{2} x^{\frac{1}{2}}$.
Using the linear approximation formula $f(a+h) \approx f(a) + h \cdot f'(a)$,where $a = 4$ and $h = -0.022$:
$f(4 - 0.022) \approx f(4) + (-0.022) \cdot f'(4)$.
$f(4) = 4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8$.
$f'(4) = \frac{3}{2} \cdot 4^{\frac{1}{2}} = \frac{3}{2} \cdot 2 = 3$.
Therefore,$f(3.978) \approx 8 + (-0.022) \cdot 3$.
$f(3.978) \approx 8 - 0.066 = 7.934$.

(D) Let $f(x) = 3^x$.
Then,the derivative is $f'(x) = 3^x \log 3$.
We need to find the value of $f(2.001)$.
Here,$a = 2$ and $h = 0.001$.
Using the linear approximation formula $f(a+h) \approx f(a) + h f'(a)$:
$f(2.001) \approx f(2) + (0.001) f'(2)$.
Substituting the values:
$f(2.001) \approx 3^2 + (0.001)(3^2 \log 3)$.
Given $\log 3 = 1.0986$:
$f(2.001) \approx 9 + (0.001)(9 \times 1.0986)$.
$f(2.001) \approx 9 + (0.001)(9.8874)$.
$f(2.001) \approx 9 + 0.0098874$.
Rounding to the given options,we get $9.00989$.