Functions, Graphs, and Limits - AP Calculus BC

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Question

Given and , what is the arc length between $0\leqt\leq t\leq4$ ?

Answer

In order to find the arc length, we must use the arc length formula for parametric curves:

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$ .

Given and , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , to derive

$\frac{dx}{dt}=(1)t^{1-1}+(0)10=1$ and

$\frac{dy}{dt}=(1)2t^{1-1}-(0)9=2$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{4}\sqrt{(1)^{2}+(2)^{2}}dt$

$L=\int_{0}^{4}(\sqrt{1+4})dt$

$L=\int_{0}^{4}(\sqrt{5})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[t\sqrt{5}]_{0}^{4}\textrm{}dt$

$L=(4)\sqrt{5}-(0)\sqrt{5}$

$L=4\sqrt{5}$

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Question

Given and , what is the length of the arc from $0\leq t\leq2$ ?

Answer

In order to find the arc length, we must use the arc length formula for parametric curves:

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$ .

Given and , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , to derive

$\frac{dx}{dt}=(1)4t^{1-1}-(0)5=4$ and

$\frac{dy}{dt}=(1)6t^{1-1}+(0)2=6$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{2}(\sqrt{(4)^{2}+(6)^{2}}dt$

$L=\int_{0}^{2}(\sqrt{16+36})dt$

$L=\int_{0}^{2}(\sqrt{52})dt$

$L=\int_{0}^{2}(\sqrt{4\times13})dt$

$L=\int_{0}^{2}(2\sqrt{13})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[2t\sqrt{13}]_{0}^{2}\textrm{}$

$L=2(2)\sqrt{13}-2(0)\sqrt{13}$

$L=4\sqrt{13}$

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Question

What is the derivative of $r=4sin\theta+2$ ?

Answer

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=4sin\theta+2$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=4cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta}{\frac{dr}{d\theta}cos\theta-rsin\theta}$

$\frac{dy}{dx}=\frac{4cos\theta\sin\theta+(4sin\theta+2)cos\theta}{4cos\theta\cos\theta-(4sin\theta+2)sin\theta}$

$\frac{dy}{dx}=\frac{4cos\theta\sin\theta+4sin\theta\cos\theta+2\cos\theta}{4cos\theta\cos\theta-4sin\theta\sin\theta-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4(cos^{2}\theta-sin^{2}\theta)-2sin\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the denominator, we get:

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4((1-\sin^{2}\theta)-sin^{2}\theta)-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4(1-2\sin^{2}\theta)-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4-8\sin^{2}\theta-2sin\theta}$

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Question

Evaluate the following limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left{\begin{matrix} x^2+1, x< 3\ x, x\geq 3 \end{matrix}\right.$

Answer

The limit we are given is one sided, meaning we are approaching our x value from one side; in this case, the negative sign exponent indicates that we are approaching 3 from the left side, or using values slightly less than three on approach.

This corresponds to the part of the piecewise function for values less than 3. When we substitute our x value being approached, we get

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Question

For the piecewise function:

$y=\left{\begin{matrix} 1 \textup{, for }x> 0\ 0 \textup{, for }x< 0\end{matrix}\right.$ , find $\lim_{x\to 0^+}$ .

Answer

The limit $\lim_{x\to 0^+}$ indicates that we are trying to find the value of the limit as approaches to zero from the right side of the graph.

From right to left approaching , the limit approaches to 1 even though the value at of the piecewise function does not exist.

The answer is .

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Question

Given the graph of above, what is $\lim_{x\rightarrow 0}f(x)$ ?

Answer

Examining the graph of the function above, we need to look at three things:

What is the limit of the function as it approaches zero from the left?

What is the limit of the function as it approaches zero from the right?

What is the function value at zero and is it equal to the first two statements?

If we look at the graph we see that as approaches zero from the left the values approach zero as well. This is also true if we look the values as approaches zero from the right. Lastly we look at the function value at zero which in this case is also zero.

Therefore, we can observe that $\lim_{x\rightarrow 0}f(x)=0$ as approaches .

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Question

Given the graph of above, what is $\lim_{x\rightarrow 0}f(x)$ ?

Answer

Examining the graph above, we need to look at three things:

What is the limit of the function as approaches zero from the left?

What is the limit of the function as approaches zero from the right?

What is the function value as and is it the same as the result from statement one and two?

Therefore, we can determine that $\lim_{x\rightarrow 0}f(x)$ does not exist, since approaches two different limits from either side : $-\infty$ from the left and $\infty$ from the right.

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Question

Given the above graph of , what is $\lim_{x\rightarrow 0^{+}}f(x)$ ?

Answer

Examining the graph, we want to find where the graph tends to as it approaches zero from the right hand side. We can see that there appears to be a vertical asymptote at zero. As the x values approach zero from the right the function values of the graph tend towards positive infinity.

Therefore, we can observe that $\lim_{x\rightarrow 0^{+}}f(x)=\infty$ as approaches from the right.

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Question

Given the above graph of , what is $\lim_{x\rightarrow 0}f(x)$ ?

Answer

Examining the graph, we can observe that $\lim_{x\rightarrow 0}f(x)$ does not exist, as is not continuous at . We can see this by checking the three conditions for which a function is continuous at a point :

A value exists in the domain of

The limit of exists as approaches

The limit of at is equal to

Given , we can see that condition #1 is not satisfied because the graph has a vertical asymptote instead of only one value for and is therefore an infinite discontinuity at .

We can also see that condition #2 is not satisfied because $\lim_{x\rightarrow 0}f(x)$ approaches two different limits: $\infty$ from the left and $-\infty$ from the right.

Based on the above, condition #3 is also not satisfied because $\lim_{x\rightarrow 0}f(x)$ is not equal to the multiple values of .

Thus, $\lim_{x\rightarrow 0}f(x)$ does not exist.

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Question

$\begin{align}&\text{What is the likely value of}\&f(-\infty)\&\text{Based on the graph?}\end{align}$

Answer

$\begin{align}&\text{Although actually observing a function at an infinite value}\&\text{of x is impossible, we can often determine via observation if}\&\text{the function approaches a certain limit. Looking at the values}\&\text{the function approaches as the absolute value of x increases}\&\text{gives a clue. Things to look out for are if the function continuously}\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\&\text{If the function continuously climbs/decreases, it likely has}\&\text{a limit that approaches positive/negative infinity. If it flattens,}\&\text{the limit is a real number. If it oscillates without control,}\&\text{the limit does not exist.}\&\text{The limit of }f(-\infty)\text{ is most likely }\text{Nonexistent.}\end{align}$

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Question

$\begin{align}&\text{What is the likely value of}\&f(-\infty)\&\text{Based on the graph?}\end{align}$

Answer

$\begin{align}&\text{Although actually observing a function at an infinite value}\&\text{of x is impossible, we can often determine via observation if}\&\text{the function approaches a certain limit. Looking at the values}\&\text{the function approaches as the absolute value of x increases}\&\text{gives a clue. Things to look out for are if the function continuously}\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\&\text{If the function continuously climbs/decreases, it likely has}\&\text{a limit that approaches positive/negative infinity. If it flattens,}\&\text{the limit is a real number. If it oscillates without control,}\&\text{the limit does not exist.}\&\text{The limit of }f(-\infty)\text{ is most likely }0\end{align}$

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Question

What is the polar form of $y=\frac{3}{9}x^{2}$ ?

Answer

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $y=\frac{3}{9}x^{2}$ , then:

$r\sin\theta=\frac{3}9{}(r\cos \theta)^{2}$

$r\sin\theta=\frac{3}{9}r^{2}\cos^{2} \theta$

Dividing both sides by $r\cos\theta$ , we get:

$\tan \theta=\frac{3}9{}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{3}{9}r$

$\tan \theta\sec\theta=\frac{3}{9}r$

$r=\frac{9}{3}\tan \theta\sec\theta$

$r=3\tan \theta\sec\theta$

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Question

$\begin{align}&\text{Which of the values most likely represents the value of f(x) at}\&x=\infty\end{align}$

Answer

$\begin{align}&\text{Although actually observing a function at an infinite value}\&\text{of x is impossible, we can often determine via observation if}\&\text{the function approaches a certain limit. Looking at the values}\&\text{the function approaches as the absolute value of x increases}\&\text{gives a clue. Things to look out for are if the function continuously}\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\&\text{If the function continuously climbs/decreases, it likely has}\&\text{a limit that approaches positive/negative infinity. If it flattens,}\&\text{the limit is a real number. If it oscillates without control,}\&\text{the limit does not exist.}\&\text{The limit of }f(\infty)\text{ is most likely }-5\end{align}$

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Question

$\begin{align}&\text{Which of the values most likely represents the value of f(x) at}\&x=\infty\end{align}$

Answer

$\begin{align}&\text{Although actually observing a function at an infinite value}\&\text{of x is impossible, we can often determine via observation if}\&\text{the function approaches a certain limit. Looking at the values}\&\text{the function approaches as the absolute value of x increases}\&\text{gives a clue. Things to look out for are if the function continuously}\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\&\text{If the function continuously climbs/decreases, it likely has}\&\text{a limit that approaches positive/negative infinity. If it flattens,}\&\text{the limit is a real number. If it oscillates without control,}\&\text{the limit does not exist.}\&\text{The limit of }f(\infty)\text{ is most likely }-\infty\end{align}$

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Question

$\begin{align}&\text{Based on the graph, decide which of the following functions values likely depicts}\&f(-\infty)\end{align}$

Answer

$\begin{align}&\text{Although actually observing a function at an infinite value}\&\text{of x is impossible, we can often determine via observation if}\&\text{the function approaches a certain limit. Looking at the values}\&\text{the function approaches as the absolute value of x increases}\&\text{gives a clue. Things to look out for are if the function continuously}\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\&\text{If the function continuously climbs/decreases, it likely has}\&\text{a limit that approaches positive/negative infinity. If it flattens,}\&\text{the limit is a real number. If it oscillates without control,}\&\text{the limit does not exist.}\&\text{The limit of }f(-\infty)\text{ is most likely }\text{Nonexistent.}\end{align}$

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Question

$\begin{align}&\text{What is the likely value of}\&f(-\infty)\&\text{Based on the graph?}\end{align}$

Answer

$\begin{align}&\text{Although actually observing a function at an infinite value}\&\text{of x is impossible, we can often determine via observation if}\&\text{the function approaches a certain limit. Looking at the values}\&\text{the function approaches as the absolute value of x increases}\&\text{gives a clue. Things to look out for are if the function continuously}\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\&\text{If the function continuously climbs/decreases, it likely has}\&\text{a limit that approaches positive/negative infinity. If it flattens,}\&\text{the limit is a real number. If it oscillates without control,}\&\text{the limit does not exist.}\&\text{The limit of }f(-\infty)\text{ is most likely }\infty\end{align}$

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Question

$\begin{align}&\text{Which of the values most likely represents the value of f(x) at}\&x=\infty\end{align}$

Answer

$\begin{align}&\text{Although actually observing a function at an infinite value}\&\text{of x is impossible, we can often determine via observation if}\&\text{the function approaches a certain limit. Looking at the values}\&\text{the function approaches as the absolute value of x increases}\&\text{gives a clue. Things to look out for are if the function continuously}\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\&\text{If the function continuously climbs/decreases, it likely has}\&\text{a limit that approaches positive/negative infinity. If it flattens,}\&\text{the limit is a real number. If it oscillates without control,}\&\text{the limit does not exist.}\&\text{The limit of }f(\infty)\text{ is most likely }\text{Nonexistent.}\end{align}$

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Question

$\begin{align}&\text{What is the likely value of}\&f(-\infty)\&\text{Based on the graph?}\end{align}$

Answer

$\begin{align}&\text{Although actually observing a function at an infinite value}\&\text{of x is impossible, we can often determine via observation if}\&\text{the function approaches a certain limit. Looking at the values}\&\text{the function approaches as the absolute value of x increases}\&\text{gives a clue. Things to look out for are if the function continuously}\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\&\text{If the function continuously climbs/decreases, it likely has}\&\text{a limit that approaches positive/negative infinity. If it flattens,}\&\text{the limit is a real number. If it oscillates without control,}\&\text{the limit does not exist.}\&\text{The limit of }f(-\infty)\text{ is most likely }\text{Nonexistent.}\end{align}$

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Question

$\begin{align}&\text{What is the likely value of}\&f(-\infty)\&\text{Based on the graph?}\end{align}$

Answer

$\begin{align}&\text{Although actually observing a function at an infinite value}\&\text{of x is impossible, we can often determine via observation if}\&\text{the function approaches a certain limit. Looking at the values}\&\text{the function approaches as the absolute value of x increases}\&\text{gives a clue. Things to look out for are if the function continuously}\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\&\text{If the function continuously climbs/decreases, it likely has}\&\text{a limit that approaches positive/negative infinity. If it flattens,}\&\text{the limit is a real number. If it oscillates without control,}\&\text{the limit does not exist.}\&\text{The limit of }f(-\infty)\text{ is most likely }0\end{align}$

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Question

$\begin{align}&\text{What is the likely value of}\&f(-\infty)\&\text{Based on the graph?}\end{align}$

Answer

$\begin{align}&\text{Although actually observing a function at an infinite value}\&\text{of x is impossible, we can often determine via observation if}\&\text{the function approaches a certain limit. Looking at the values}\&\text{the function approaches as the absolute value of x increases}\&\text{gives a clue. Things to look out for are if the function continuously}\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\&\text{If the function continuously climbs/decreases, it likely has}\&\text{a limit that approaches positive/negative infinity. If it flattens,}\&\text{the limit is a real number. If it oscillates without control,}\&\text{the limit does not exist.}\&\text{The limit of }f(-\infty)\text{ is most likely }\frac{1}{4}\end{align}$

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