Final, Fall 2023

We are trying to evaluate:

$$\underset{x\to 0^{+}}{\lim }\frac{\sin (x)(e^x-1)}{x^3}$$
\lim_{x \to 0^+} \frac{\sin(x)(e^x - 1)}{x^3}


We can rewrite this as:

$$\underset{x\to 0^{+}}{\lim }(\frac{\sin (x)}{x}\cdot \frac{e^x-1}{x}\cdot \frac{1}{x})$$
\lim_{x \to 0^+} \left( \frac{\sin(x)}{x} \cdot \frac{e^x - 1}{x} \cdot \frac{1}{x} \right)


By splitting it like this, we can analyze each part separately, making it easier to solve.



1. **Evaluate** $\lim_{x \to 0^+} \frac{\sin(x)}{x}$:

$$\underset{x\to 0^{+}}{\lim }\frac{\sin (x)}{x}=1$$
\lim_{x \to 0^+} \frac{\sin(x)}{x} = 1


This is a standard limit that often appears in exams.

[Remember this limit: $$\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=1$$
\lim_{x \to 0} \frac{\sin(x)}{x} = 1

They love to test this! It shows that $\sin(x)$ and $x$ are almost identical for small values of $x$.]

2. **Evaluate** $\lim_{x \to 0^+} \frac{e^x - 1}{x}$:

$$\underset{x\to 0^{+}}{\lim }\frac{e^x-1}{x}=1$$
\lim_{x \to 0^+} \frac{e^x - 1}{x} = 1


For small $x$, the function $e^x - 1$ behaves similarly to $x$.

[This limit is another must-know: $$\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$$
\lim_{x \to 0} \frac{e^x - 1}{x} = 1

This shows that for small $x$, the numerator and denominator act like the same function.]

3. **Evaluate** $\lim_{x \to 0^+} \frac{1}{x}$:

$$\underset{x\to 0^{+}}{\lim }\frac{1}{x}=\mathrm{\infty }$$
\lim_{x \to 0^+} \frac{1}{x} = \infty


This tells us that the function grows without bound as $x$ approaches 0 from the positive side.



Using the property that:

$$\underset{x\to 0^{+}}{\lim }(f(x)\cdot g(x)\cdot h(x))=\underset{x\to 0^{+}}{\lim }f(x)\cdot \underset{x\to 0^{+}}{\lim }g(x)\cdot \underset{x\to 0^{+}}{\lim }h(x)$$
\lim_{x \to 0^+} \left( f(x) \cdot g(x) \cdot h(x) \right) = \lim_{x \to 0^+} f(x) \cdot \lim_{x \to 0^+} g(x) \cdot \lim_{x \to 0^+} h(x)


**(only if all limits are finite or one limit goes to infinity without any undefined behavior like $0 \cdot \infty$)**, we get:

$$1\cdot 1\cdot \mathrm{\infty }=\mathrm{\infty }$$
1 \cdot 1 \cdot \infty = \infty


Since the multiplication involves finite values and one $\infty$, the result is:

$$\boxed{{\displaystyle \mathrm{\infty }}}$$
\boxed{\infty}


This approach works here because each limit we evaluated behaves predictably and does not lead to an undefined form like $0 \cdot \infty$. Therefore, we can confidently multiply the individual limits.


In this tough question they are testing your knowledge of the these two limits:
$$\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=1$$
\lim_{x \to 0} \frac{\sin(x)}{x} = 1
and

$$\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$$
\lim_{x \to 0} \frac{e^x - 1}{x} = 1
. As long as you know those limits then you'll have no problem with this question.


The reciprocal of each limit is also 1:

$$\underset{x\to 0}{\lim }\frac{x}{\sin (x)}=1\text{and}\underset{x\to 0}{\lim }\frac{x}{e^x-1}=1$$
\lim_{x \to 0} \frac{x}{\sin(x)} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{x}{e^x - 1} = 1


These limits show that $\sin(x)$ and $e^x - 1$ behave almost exactly like $x$ when $x$ is close to 0. Knowing these will help you solve many tricky limits that appear in calculus exams!



You can also therefore mix these terms:

$$\underset{x\to 0}{\lim }\frac{\sin (x)}{e^x-1}=1$$
\lim_{x \to 0} \frac{\sin(x)}{e^x - 1} = 1


Both $\sin(x)$ and $e^x - 1$ behave like $x$ when $x$ is close to 0, so this limit simplifies to:

$$\underset{x\to 0}{\lim }\frac{x}{x}=1$$
\lim_{x \to 0} \frac{x}{x} = 1


This is another handy limit to keep in mind for calculus exams!

We can multiply and divide the expression by its conjugate:

$$\underset{x\to \mathrm{\infty }}{\lim }(\sqrt{4+x^2}-x)\cdot \frac{\sqrt{4+x^2}+x}{\sqrt{4+x^2}+x}$$
\lim_{x \to \infty} \left( \sqrt{4 + x^2} - x \right) \cdot \frac{\sqrt{4 + x^2} + x}{\sqrt{4 + x^2} + x}


This simplifies to:

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{(4+x^2)-x^2}{\sqrt{4+x^2}+x}$$
\lim_{x \to \infty} \frac{(4 + x^2) - x^2}{\sqrt{4 + x^2} + x}




$$(4+x^2)-x^2=4$$
(4 + x^2) - x^2 = 4


So the expression becomes:

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{4}{\sqrt{4+x^2}+x}$$
\lim_{x \to \infty} \frac{4}{\sqrt{4 + x^2} + x}




As $x \to \infty$, the term $\sqrt{4 + x^2}$ behaves similarly to $x$ because $x^2$ dominates inside the square root. Thus:

$$\sqrt{4+x^2}\approx x$$
\sqrt{4 + x^2} \approx x


So the denominator becomes approximately:

$$\sqrt{4+x^2}+x\approx x+x=2x$$
\sqrt{4 + x^2} + x \approx x + x = 2x


Now, the expression becomes:

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{4}{2x}=\underset{x\to \mathrm{\infty }}{\lim }\frac{2}{x}=0$$
\lim_{x \to \infty} \frac{4}{2x} = \lim_{x \to \infty} \frac{2}{x} = 0




$$\boxed{{\displaystyle 0}}$$
\boxed{0}


Thus, the correct answer is:

$$\boxed{{\displaystyle (D)\hspace{0.17em}0}}$$
\boxed{(D) \, 0}

We recognize that the integral of $\frac{1}{1 + x^2}$ is the well-known inverse tangent function:

$$\int \frac{1}{1+x^2}\hspace{0.17em}dx=\arctan (x)+C_1$$
\int \frac{1}{1 + x^2} \, dx = \arctan(x) + C_1




The integral of $\sin(x)$ is:

$$\int \sin (x)\hspace{0.17em}dx=-\cos (x)+C_2$$
\int \sin(x) \, dx = -\cos(x) + C_2




Now we combine the two integrals:

$$\int (\frac{1}{1+x^2}+\sin (x))\hspace{0.17em}dx=\arctan (x)-\cos (x)+C$$
\int \left( \frac{1}{1 + x^2} + \sin(x) \right) \, dx = \arctan(x) - \cos(x) + C


where $C = C_1 + C_2$ is the most general constant of integration.



Thus, the most general antiderivative is:

$$\boxed{{\displaystyle \arctan (x)-\cos (x)+C}}$$
\boxed{\arctan(x) - \cos(x) + C}


The correct answer is:

$$\boxed{{\displaystyle (A)\hspace{0.17em}\arctan (x)-\cos (x)+C}}$$
\boxed{(A) \, \arctan(x) - \cos(x) + C}

Using the product rule on the left side:

$$\frac{d}{dx}((1+x+x^2)e^p)=\frac{d}{dx}(1)$$
\frac{d}{dx} \left( (1 + x + x^2) e^p \right) = \frac{d}{dx}(1)


The derivative of the right side is:

$$\frac{d}{dx}(1)=0$$
\frac{d}{dx}(1) = 0


Now apply the product rule to the left side:

$$\frac{d}{dx}(1+x+x^2)\cdot e^p+(1+x+x^2)\cdot \frac{d}{dx}(e^p)=0$$
\frac{d}{dx} \left( 1 + x + x^2 \right) \cdot e^p + (1 + x + x^2) \cdot \frac{d}{dx}(e^p) = 0


This simplifies to:

$$(1+2x)e^p+(1+x+x^2)e^p\frac{dp}{dx}=0$$
(1 + 2x) e^p + (1 + x + x^2) e^p \frac{dp}{dx} = 0




First, isolate the term with $\frac{dp}{dx}$:

$$(1+x+x^2)e^p\frac{dp}{dx}=-(1+2x)e^p$$
(1 + x + x^2) e^p \frac{dp}{dx} = -(1 + 2x) e^p


Now, divide both sides by $(1 + x + x^2) e^p$ to get:

$$\frac{dp}{dx}=-\frac{1+2x}{1+x+x^2}$$
\frac{dp}{dx} = -\frac{1 + 2x}{1 + x + x^2}




When $x = 1$, the expression becomes:

$$\frac{dp}{dx}=-\frac{1+2(1)}{1+1+1^2}=-\frac{1+2}{1+1+1}=-\frac{3}{3}=-1$$
\frac{dp}{dx} = -\frac{1 + 2(1)}{1 + 1 + 1^2} = -\frac{1 + 2}{1 + 1 + 1} = -\frac{3}{3} = -1




Thus, the rate of change $\frac{dp}{dx}$ when $x = 1$ is:

$$\boxed{{\displaystyle -1}}$$
\boxed{-1}


The correct answer is:

$$\boxed{{\displaystyle (A)\hspace{0.17em}-1}}$$
\boxed{(A) \, -1}

The distance $d$ from any point $(x, y)$ on the parabola to the origin is:

$$d=\sqrt{x^2+y^2}$$
d = \sqrt{x^2 + y^2}


We want to minimize this distance. Since minimizing the square of the distance is easier, we’ll minimize:

$$d^2=x^2+y^2$$
d^2 = x^2 + y^2




$$d^2=(4-y^2{)}^2+y^2$$
d^2 = (4 - y^2)^2 + y^2




$$d^2=16-8y^2+y^4+y^2=y^4-7y^2+16$$
d^2 = 16 - 8y^2 + y^4 + y^2 = y^4 - 7y^2 + 16




Take the derivative of $y^4 - 7y^2 + 16$ and set it to 0:

$$\frac{d}{dy}(y^4-7y^2+16)=4y^3-14y$$
\frac{d}{dy}(y^4 - 7y^2 + 16) = 4y^3 - 14y


Set the derivative equal to 0:

$$4y^3-14y=0$$
4y^3 - 14y = 0


Factor out $2y$:

$$2y(2y^2-7)=0$$
2y(2y^2 - 7) = 0


Solutions:

$$y=0\text{or}y=\pm \sqrt{\frac{7}{2}}$$
y = 0 \quad \text{or} \quad y = \pm \sqrt{\frac{7}{2}}




For $y = 0$:

$$x=4$$
x = 4


For $y = \pm \sqrt{7/2}$:

$$x=4-\frac{7}{2}=\frac{1}{2}$$
x = 4 - \frac{7}{2} = \frac{1}{2}




The $x$-coordinates closest to the origin are:

$$x=4\text{and}x=\frac{1}{2}$$
x = 4 \quad \text{and} \quad x = \frac{1}{2}


Thus, the correct answer is:

$$\boxed{{\displaystyle (A)\hspace{0.17em}\frac{1}{2}}}$$
\boxed{(A) \, \frac{1}{2}}

We are given a right-angled triangle with a hypotenuse that is always 5 cm long. The sides are $x$, $y$, and the hypotenuse is $5$. So the Pythagorean theorem gives:

$$x^2+y^2=5^2=25$$
x^2 + y^2 = 5^2 = 25




$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0


Simplify:

$$x\frac{dx}{dt}+y\frac{dy}{dt}=0$$
x \frac{dx}{dt} + y \frac{dy}{dt} = 0




$$y\frac{dy}{dt}=-x\frac{dx}{dt}$$
y \frac{dy}{dt} = -x \frac{dx}{dt}


Divide both sides by $x' \neq 0$:

$$\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$$
\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}


Thus:

$$\frac{y^{\mathrm{\prime }}}{x^{\mathrm{\prime }}}=-\frac{x}{y}$$
\frac{y'}{x'} = -\frac{x}{y}




Using the Pythagorean theorem:

$$4^2+y^2=25\Rightarrow 16+y^2=25\Rightarrow y^2=9\Rightarrow y=3$$
4^2 + y^2 = 25 \quad \Rightarrow \quad 16 + y^2 = 25 \quad \Rightarrow \quad y^2 = 9 \quad \Rightarrow \quad y = 3




$$\frac{y^{\mathrm{\prime }}}{x^{\mathrm{\prime }}}=-\frac{x}{y}=-\frac{4}{3}$$
\frac{y'}{x'} = -\frac{x}{y} = -\frac{4}{3}




The ratio $\frac{y'}{x'}$ when $x = 4$ is:

$$\boxed{{\displaystyle -\frac{4}{3}}}$$
\boxed{-\frac{4}{3}}


Thus, the correct answer is:

$$\boxed{{\displaystyle (A)\hspace{0.17em}-\frac{4}{3}}}$$
\boxed{(A) \, -\frac{4}{3}}

We use logarithmic differentiation. Take the natural logarithm of both sides:

$$\ln (f(x))=\cosh (x)\ln (x+2)$$
\ln(f(x)) = \cosh(x) \ln(x + 2)




Using the chain rule on the left side, the derivative of $\ln(f(x))$ is:

$$\frac{d}{dx}(\ln (f(x)))=\frac{f^{\mathrm{\prime }}(x)}{f(x)}$$
\frac{d}{dx} \left( \ln(f(x)) \right) = \frac{f'(x)}{f(x)}


This term $\frac{f'(x)}{f(x)}$ comes from the **chain rule**. When differentiating $\ln(f(x))$, we first take $\frac{1}{f(x)}$ and then multiply by the derivative of the inner function $f(x)$, which gives:

$$\frac{d}{dx}(\ln (f(x)))=\frac{f^{\mathrm{\prime }}(x)}{f(x)}$$
\frac{d}{dx} \left( \ln(f(x)) \right) = \frac{f'(x)}{f(x)}


On the right side, we apply the **product rule**:

$$\frac{f^{\mathrm{\prime }}(x)}{f(x)}=\sinh (x)\ln (x+2)+\cosh (x)\cdot \frac{1}{x+2}$$
\frac{f'(x)}{f(x)} = \sinh(x) \ln(x + 2) + \cosh(x) \cdot \frac{1}{x + 2}




We need to find $f'(0)$. Start by plugging in $x = 0$ into the right side:

$$\sinh (0)\ln (2)+\cosh (0)\cdot \frac{1}{2}$$
\sinh(0) \ln(2) + \cosh(0) \cdot \frac{1}{2}


Since:

$$\sinh (0)=0\text{and}\cosh (0)=1$$
\sinh(0) = 0 \quad \text{and} \quad \cosh(0) = 1


This simplifies to:

$$0\cdot \ln (2)+1\cdot \frac{1}{2}=\frac{1}{2}$$
0 \cdot \ln(2) + 1 \cdot \frac{1}{2} = \frac{1}{2}




Now, multiply both sides by $f(0)$. At $x = 0$, we have:

$$f(0)=(0+2{)}^{\cosh (0)}=2^1=2$$
f(0) = (0 + 2)^{\cosh(0)} = 2^1 = 2


Thus:

$$f^{\mathrm{\prime }}(0)=2\cdot \frac{1}{2}=1$$
f'(0) = 2 \cdot \frac{1}{2} = 1




The value of $f'(0)$ is:

$$\boxed{{\displaystyle 1}}$$
\boxed{1}


Thus, the correct answer is:

$$\boxed{{\displaystyle (B)\hspace{0.17em}1}}$$
\boxed{(B) \, 1}

We use the quotient rule to differentiate:

$$f(x)=\frac{x^2}{1+x^2}$$
f(x) = \frac{x^2}{1 + x^2}


The quotient rule states:

$$\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right)=\frac{u^{\mathrm{\prime }}(x)v(x)-u(x)v^{\mathrm{\prime }}(x)}{(v(x){)}^2}$$
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}


where $u(x) = x^2$ and $v(x) = 1 + x^2$.



$$u^{\mathrm{\prime }}(x)=2x,v^{\mathrm{\prime }}(x)=2x$$
u'(x) = 2x, \quad v'(x) = 2x


Now:

$$f^{\mathrm{\prime }}(x)=\frac{(2x)(1+x^2)-(x^2)(2x)}{(1+x^2{)}^2}$$
f'(x) = \frac{(2x)(1 + x^2) - (x^2)(2x)}{(1 + x^2)^2}


Simplify:

$$(2x)(1+x^2)-(x^2)(2x)=2x+2x^3-2x^3=2x$$
(2x)(1 + x^2) - (x^2)(2x) = 2x + 2x^3 - 2x^3 = 2x


Thus:

$$f^{\mathrm{\prime }}(x)=\frac{2x}{(1+x^2{)}^2}$$
f'(x) = \frac{2x}{(1 + x^2)^2}




$$f^{\mathrm{\prime }}(1)=\frac{2(1)}{(1+1^2{)}^2}=\frac{2}{(1+1{)}^2}=\frac{2}{4}=\frac{1}{2}$$
f'(1) = \frac{2(1)}{(1 + 1^2)^2} = \frac{2}{(1 + 1)^2} = \frac{2}{4} = \frac{1}{2}




The slope of the tangent line at $x = 1$ is:

$$\boxed{{\displaystyle \frac{1}{2}}}$$
\boxed{\frac{1}{2}}


Thus, the correct answer is:

$$\boxed{{\displaystyle (B)\hspace{0.17em}\frac{1}{2}}}$$
\boxed{(B) \, \frac{1}{2}}

We already know the derivative is:

$$f^{\mathrm{\prime }}(x)=\frac{2x}{(1+x^2{)}^2}$$
f'(x) = \frac{2x}{(1 + x^2)^2}


Evaluating it at $x = 1$:

$$f^{\mathrm{\prime }}(1)=\frac{2(1)}{(1+1^2{)}^2}=\frac{2}{4}=\frac{1}{2}$$
f'(1) = \frac{2(1)}{(1 + 1^2)^2} = \frac{2}{4} = \frac{1}{2}


So:

$$m=\frac{1}{2}$$
m = \frac{1}{2}




$$f(1)=\frac{1^2}{1+1^2}=\frac{1}{2}$$
f(1) = \frac{1^2}{1 + 1^2} = \frac{1}{2}


Thus, the point is:

$$(1,\frac{1}{2})$$
(1, \frac{1}{2})




The equation of the line is:

$$y=mx+b$$
y = mx + b


Substitute $m = \frac{1}{2}$ and $(1, \frac{1}{2})$:

$$\frac{1}{2}=\frac{1}{2}(1)+b$$
\frac{1}{2} = \frac{1}{2}(1) + b


Simplify:

$$\frac{1}{2}=\frac{1}{2}+b$$
\frac{1}{2} = \frac{1}{2} + b


Solve for $b$:

$$b=0$$
b = 0




The $y$-intercept is:

$$\boxed{{\displaystyle 0}}$$
\boxed{0}


Thus, the correct answer is:

$$\boxed{{\displaystyle (A)\hspace{0.17em}0}}$$
\boxed{(A) \, 0}

$$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\dots$$
\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots


Since higher-order terms become negligibly small for small $x$, we approximate:

$$\ln (1+x)\approx x-\frac{x^2}{2}$$
\ln(1 + x) \approx x - \frac{x^2}{2}




$$\frac{1}{\ln (1+x)}=\frac{1}{x-\frac{x^2}{2}}$$
\frac{1}{\ln(1 + x)} = \frac{1}{x - \frac{x^2}{2}}


Factor out $x$:

$$\frac{1}{\ln (1+x)}=\frac{1}{x(1-\frac{x}{2})}$$
\frac{1}{\ln(1 + x)} = \frac{1}{x \left( 1 - \frac{x}{2} \right)}


**Remember:** The approximation $\frac{1}{1 - u} \approx 1 + u$ is useful for small $u$.

Using this approximation:

$$\frac{1}{\ln (1+x)}\approx \frac{1}{x}+\frac{1}{2}$$
\frac{1}{\ln(1 + x)} \approx \frac{1}{x} + \frac{1}{2}




$$\underset{x\to 0}{\lim }((1+\frac{1}{x})-(\frac{1}{x}+\frac{1}{2}))$$
\lim_{x \to 0} \left( \left( 1 + \frac{1}{x} \right) - \left( \frac{1}{x} + \frac{1}{2} \right) \right)


Simplify:

$$\underset{x\to 0}{\lim }(1+\frac{1}{x}-\frac{1}{x}-\frac{1}{2})=\underset{x\to 0}{\lim }(1-\frac{1}{2})$$
\lim_{x \to 0} \left( 1 + \frac{1}{x} - \frac{1}{x} - \frac{1}{2} \right) = \lim_{x \to 0} \left( 1 - \frac{1}{2} \right)


$$\underset{x\to 0}{\lim }\frac{1}{2}=\frac{1}{2}$$
\lim_{x \to 0} \frac{1}{2} = \frac{1}{2}




The value of the limit is:

$$\boxed{{\displaystyle \frac{1}{2}}}$$
\boxed{\frac{1}{2}}


Thus, the correct answer is:

$$\boxed{{\displaystyle (D)\hspace{0.17em}\frac{1}{2}}}$$
\boxed{(D) \, \frac{1}{2}}




This question is trying to teach you something about the comparative size of these two functions as we get closer to 0.

It turns out that $\frac{x + 1}{x}$ is about $1/2$ bigger than $\frac{1}{\ln(1 + x)}$ as we get close to 0, even though they both approach $\infty$. That's what this question is trying to teach you!

A function $f(x)$ is continuous at $x = 0$ if:

$$\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }f(x)=f(0)$$
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)




$$f(0)=\cos (0-\pi k)=\cos (\pi k)$$
f(0) = \cos(0 - \pi k) = \cos(\pi k)




$$\underset{x\to 0^{-}}{\lim }f(x)=\cos (\pi k)$$
\lim_{x \to 0^-} f(x) = \cos(\pi k)




Using Taylor expansions:

$$\sin (x)=x-\frac{x^3}{6}+\dots ,\cos (x)=1-\frac{x^2}{2}+\dots$$
\sin(x) = x - \frac{x^3}{6} + \dots, \quad \cos(x) = 1 - \frac{x^2}{2} + \dots


$$\frac{\sin (x)-x}{1-\cos (x)}=\frac{-\frac{x^3}{6}}{\frac{x^2}{2}}=\frac{-x}{3}$$
\frac{\sin(x) - x}{1 - \cos(x)} = \frac{-\frac{x^3}{6}}{\frac{x^2}{2}} = \frac{-x}{3}


Thus:

$$\underset{x\to 0^{+}}{\lim }f(x)=0$$
\lim_{x \to 0^+} f(x) = 0




$$\cos (\pi k)=0$$
\cos(\pi k) = 0




$$\pi k=\frac{\pi }{2},\frac{3\pi }{2},\dots \Rightarrow k=\frac{1}{2},\frac{3}{2},\dots$$
\pi k = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \quad \Rightarrow \quad k = \frac{1}{2}, \frac{3}{2}, \dots


The smallest positive value is:

$$k=\frac{1}{2}$$
k = \frac{1}{2}




$$\boxed{{\displaystyle (B)\hspace{0.17em}k=\frac{1}{2}}}$$
\boxed{(B) \, k = \frac{1}{2}}

The parabola is given by the equation:

$$y=(x-r_1)(x-r_2)\mathrm{.}$$
y = (x - r_1)(x - r_2).


The roots are $r_1$ and $r_2$, and the **distance between the roots** is:

$$d=r_2-r_1\mathrm{.}$$
d = r_2 - r_1.


The vertex of the parabola lies at the midpoint of the roots. The x-coordinate of the vertex is:

$$h=\frac{r_1+r_2}{2}\mathrm{.}$$
h = \frac{r_1 + r_2}{2}.


The **y-coordinate of the vertex** is given by:

$$y_{\text{vertex}}=-\frac{d^2}{4}\mathrm{.}$$
y_{\text{vertex}} = -\frac{d^2}{4}.


Since the equation has a negative coefficient for the quadratic term, the parabola opens downward, meaning the vertex represents a maximum point.



To find how the $y$-coordinate of the vertex changes with time, we differentiate the formula for $y_{\text{vertex}}$ with respect to time $t$:

$$\frac{d{y}_{\text{vertex}}}{dt}=-\frac{1}{4}\cdot 2d\cdot \frac{dd}{dt}\mathrm{.}$$
\frac{dy_{\text{vertex}}}{dt} = -\frac{1}{4} \cdot 2d \cdot \frac{dd}{dt}.


Simplify the expression:

$$\frac{d{y}_{\text{vertex}}}{dt}=-\frac{d}{2}\cdot \frac{dd}{dt}\mathrm{.}$$
\frac{dy_{\text{vertex}}}{dt} = -\frac{d}{2} \cdot \frac{dd}{dt}.




Given that:
$$d=2,\frac{dd}{dt}=1\hspace{0.17em}\text{unit per second},$$
d = 2, \quad \frac{dd}{dt} = 1 \, \text{unit per second},


substitute these values into the equation:

$$\frac{d{y}_{\text{vertex}}}{dt}=-\frac{2}{2}\cdot 1=-1$$
\frac{dy_{\text{vertex}}}{dt} = -\frac{2}{2} \cdot 1 = -1




Thus, the rate of change of the $y$-coordinate of the vertex is:

$$\boxed{{\displaystyle -1}}\hspace{0.17em}\text{unit per second}\mathrm{.}$$
\boxed{-1} \, \text{unit per second}.


The negative sign means the vertex is moving downward as the roots move apart.

The volume is:

$$V=\pi r^{2}h=1000\text{(Equation 1)}$$
V = \pi r^2 h = 1000 \quad \text{(Equation 1)}




The surface area is:

$$A=\pi r^2+2\pi rh\text{(Equation 2)}$$
A = \pi r^2 + 2\pi r h \quad \text{(Equation 2)}




Solve for $h$:

$$h=\frac{1000}{\pi r^2}$$
h = \frac{1000}{\pi r^2}


Substitute into the surface area formula:

$$A=\pi r^2+\frac{2000}{r}\mathrm{.}$$
A = \pi r^2 + \frac{2000}{r}.




Take the derivative:

$$\frac{dA}{dr}=2\pi r-\frac{2000}{r^2}$$
\frac{dA}{dr} = 2\pi r - \frac{2000}{r^2}


Set the derivative equal to 0:

$$2\pi r=\frac{2000}{r^2}\Rightarrow r^3=\frac{1000}{\pi }\mathrm{.}$$
2\pi r = \frac{2000}{r^2} \quad \Rightarrow \quad r^3 = \frac{1000}{\pi}.


Thus:

$$r=\frac{10}{\sqrt[3]{\pi }}\mathrm{.}$$
r = \frac{10}{\sqrt[3]{\pi}}.




$$h=\frac{1000}{\pi r^2}\mathrm{.}$$
h = \frac{1000}{\pi r^2}.


Since $r = \frac{10}{\pi^{1/3}}$, we substitute:

$$h=\frac{1000}{\pi {\left(\frac{10}{{\pi }^{1\mathrm{/}3}}\right)}^2}\mathrm{.}$$
h = \frac{1000}{\pi \left( \frac{10}{\pi^{1/3}} \right)^2}.


$${\left(\frac{10}{{\pi }^{1\mathrm{/}3}}\right)}^2=\frac{100}{{\pi }^{2\mathrm{/}3}}\mathrm{.}$$
\left( \frac{10}{\pi^{1/3}} \right)^2 = \frac{100}{\pi^{2/3}}.


$$h=\frac{1000}{\pi \cdot \frac{100}{{\pi }^{2\mathrm{/}3}}}=\frac{1000\cdot {\pi }^{2\mathrm{/}3}}{100\cdot \pi }=\frac{10\cdot {\pi }^{2\mathrm{/}3}}{\pi }\mathrm{.}$$
h = \frac{1000}{\pi \cdot \frac{100}{\pi^{2/3}}} = \frac{1000 \cdot \pi^{2/3}}{100 \cdot \pi} = \frac{10 \cdot \pi^{2/3}}{\pi}.


$$h=10\cdot {\pi }^{-1\mathrm{/}3}=\frac{10}{{\pi }^{1\mathrm{/}3}}\mathrm{.}$$
h = 10 \cdot \pi^{-1/3} = \frac{10}{\pi^{1/3}}.




To confirm that this is a minimum, we take the second derivative:

$$\frac{dA}{dr}=2\pi r-\frac{2000}{r^2},$$
\frac{dA}{dr} = 2\pi r - \frac{2000}{r^2},


$$\frac{d^{2}A}{d{r}^2}=2\pi +\frac{4000}{r^3}\mathrm{.}$$
\frac{d^2A}{dr^2} = 2\pi + \frac{4000}{r^3}.


Since $r > 0$, both terms are positive:

$$\frac{d^{2}A}{d{r}^2}>0.$$
\frac{d^2A}{dr^2} > 0.


This confirms the solution corresponds to a minimum.



The dimensions that minimize the material used are:

$$\boxed{{\displaystyle r=\frac{10}{{\pi }^{1\mathrm{/}3}},h=\frac{10}{{\pi }^{1\mathrm{/}3}}\mathrm{.}}}$$
\boxed{r = \frac{10}{\pi^{1/3}}, \quad h = \frac{10}{\pi^{1/3}}.}