Derivative at a point

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AP Calculus AB › Derivative at a point

Questions 1 - 10

$\begin{align*}&\text{Approximate the acceleration of a particle using}\&\text{a central difference approach at }t=6\&\text{Using the following table of positions:}\&t&p(t)\&2&26\&6&-23\&10&18\&14&16\&18&-34\end{align*}$

$\frac{45}{8}$

$-\frac{49}{4}$

$-\frac{1}{2}$

Explanation

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\&\text{words, you take the derivative of a position function twice}\&\text{to define it. In this problem, we’re not given a function, but}\&\text{ a table of times and positions, requiring an approximation,}\&\text{a 2nd order central difference. This has the form:}\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{This approximation is actually born from two types of first}\&\text{order approximations: the forward and backward differences,}\&\text{which have the forms:}\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\&\text{The difference of these two, divided by our increment, gives}\&\text{a derivative of a derivative.}\&\text{Applying the formula above, with }h=4\text{, we find:}\&a(6)\approx\frac{18+(26)-2(-23)}{4^2}\&a(6)\approx\frac{45}{8}\end{align*}$

$\begin{align*}&x&f(x)\&3&27\&8&8\&13&43\&18&8\&23&-49\&28&-38\&33&37\&\text{Approximate a backward derivative at }x=33\end{align*}$

$-\frac{37}{5}$

$\frac{37}{5}$

Explanation

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\&\text{two values: one at the point of interest and another preceding}\&\text{it, and finally that distance between the two. Mathematically,}\&\text{this follows the form:}\&f'(x)=\frac{f(x)-f(x-h)}{h}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{In approximating a derivative, it is generally prudent to minimize}\&\text{the distance between function values. In other words, it’s a}\&\text{good idea to look at two adjacent points. For our problem, that’d}\&\text{be:}\&f(33)\text{ and }f(28)\&\text{Applying the formula above, with }h=5\text{, we find:}\&f'(33)\approx\frac{37-(-38)}{5}\&f'(33)\approx15\end{align*}$

$\begin{align*}&\text{Approximate a forward derivative at }x=4\&\text{Using the following table:}\&x&&f(x)\&1&&5\&4&&-8\&7&&15\end{align*}$

$\frac{23}{3}$

$-\frac{13}{3}$

$\frac{23}{6}$

$-\frac{8}{3}$

Explanation

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\&\text{two values: one at the point of interest and another succeeding}\&\text{it, and finally that distance between the two. Mathematically,}\&\text{this follows the form:}\&f'(x)=\frac{f(x+h)-f(x)}{h}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{In approximating a derivative, it is generally prudent to minimize}\&\text{the distance between function values. In other words, it’s a}\&\text{good idea to look at two adjacent points. For our problem, that’d}\&\text{be:}\&f(4)\text{ and }f(7)\&\text{Applying the formula above, with }h=3\text{, we find:}\&f'(4)\approx\frac{15-(15)}{3}\&f'(4)\approx\frac{23}{3}\end{align*}$

$\begin{align*}&t&p(t)\&2&24\&3&6\&4&-32\&5&10\&6&-20\&7&-37\&8&-29\&9&40\&\text{The above table lists a particle's postion at different times.}\&\text{Approximate its velocity with a forward difference at }t=2\end{align*}$

Explanation

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\&\text{we do not have a position function, we cannot take the derivative;}\&\text{we must use an approximation. To find a forward derivative approximation,}\&\text{we consider function two values: one at the point of interest}\&\text{and another succeeding it, and finally that distance between}\&\text{the two. Mathematically, this follows the form:}\&f'(x)=\frac{f(x+h)-f(x)}{h}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{In approximating a derivative, it is generally prudent to minimize}\&\text{the distance between function values. In other words, it’s a}\&\text{good idea to look at two adjacent points. For our problem, that’d}\&\text{be:}\&f(2)\text{ and }f(3)\&\text{Applying the formula above, with }h=1\text{, we find:}\&v(2)\approx\frac{6-(24)}{1}\&v(2)\approx-18\end{align*}$

$\begin{align*}&x&f(x)\&-3&12\&1&-14\&5&-45\&9&-1\&13&-31\&17&-38\&\text{Approximate a backward derivative at }x=1\end{align*}$

$-\frac{13}{2}$

$\frac{13}{2}$

$\frac{1}{2}$

$-\frac{1}{2}$

Explanation

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\&\text{two values: one at the point of interest and another preceding}\&\text{it, and finally that distance between the two. Mathematically,}\&\text{this follows the form:}\&f'(x)=\frac{f(x)-f(x-h)}{h}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{In approximating a derivative, it is generally prudent to minimize}\&\text{the distance between function values. In other words, it’s a}\&\text{good idea to look at two adjacent points. For our problem, that’d}\&\text{be:}\&f(1)\text{ and }f(-3)\&\text{Applying the formula above, with }h=4\text{, we find:}\&f'(1)\approx\frac{-14-(12)}{4}\&f'(1)\approx-\frac{13}{2}\end{align*}$

$\begin{align*}&t&p(t)\&3&-31\&4&-41\&5&-19\&6&-4\&7&-40\&\text{The above table lists a particle's postion at different times.}\&\text{Approximate its velocity with a forward difference at }t=6\end{align*}$

Explanation

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\&\text{we do not have a position function, we cannot take the derivative;}\&\text{we must use an approximation. To find a forward derivative approximation,}\&\text{we consider function two values: one at the point of interest}\&\text{and another succeeding it, and finally that distance between}\&\text{the two. Mathematically, this follows the form:}\&f'(x)=\frac{f(x+h)-f(x)}{h}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{In approximating a derivative, it is generally prudent to minimize}\&\text{the distance between function values. In other words, it’s a}\&\text{good idea to look at two adjacent points. For our problem, that’d}\&\text{be:}\&f(6)\text{ and }f(7)\&\text{Applying the formula above, with }h=1\text{, we find:}\&v(6)\approx\frac{-40-(-4)}{1}\&v(6)\approx-36\end{align*}$

$\begin{align*}&t&p(t)\&2&-28\&4&25\&6&-25\&8&1\&10&20\&12&39\&14&46\&16&5\&\text{The table above lists the positions of a particle}\&\text{At various times. Approximate the acceleration at }t=4\&\text{Using a central difference approach.}\end{align*}$

$-\frac{103}{4}$

$\frac{53}{2}$

$\frac{3}{4}$

$\frac{3}{2}$

Explanation

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\&\text{words, you take the derivative of a position function twice}\&\text{to define it. In this problem, we’re not given a function, but}\&\text{ a table of times and positions, requiring an approximation,}\&\text{a 2nd order central difference. This has the form:}\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{This approximation is actually born from two types of first}\&\text{order approximations: the forward and backward differences,}\&\text{which have the forms:}\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\&\text{The difference of these two, divided by our increment, gives}\&\text{a derivative of a derivative.}\&\text{Applying the formula above, with }h=2\text{, we find:}\&a(4)\approx\frac{-25+(-28)-2(25)}{2^2}\&a(4)\approx-\frac{103}{4}\end{align*}$

$\begin{align*}&x&f(x)\&1&-27\&3&-38\&5&11\&7&-5\&9&-4\&11&16\&13&27\&15&-15\&\text{Using the above table, make a forward derivative approximation at }x=9\end{align*}$

$\frac{1}{2}$

Explanation

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\&\text{two values: one at the point of interest and another succeeding}\&\text{it, and finally that distance between the two. Mathematically,}\&\text{this follows the form:}\&f'(x)=\frac{f(x+h)-f(x)}{h}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{In approximating a derivative, it is generally prudent to minimize}\&\text{the distance between function values. In other words, it’s a}\&\text{good idea to look at two adjacent points. For our problem, that’d}\&\text{be:}\&f(9)\text{ and }f(11)\&\text{Applying the formula above, with }h=2\text{, we find:}\&f'(9)\approx\frac{16-(16)}{2}\&f'(9)\approx10\end{align*}$

$\begin{align*}&\text{Approximate a forward derivative at }x=7\&\text{Using the following table:}\&x&f(x)\&1&8\&3&4\&5&37\&7&-24\&9&-18\&11&-38\&13&44\end{align*}$

$-\frac{61}{2}$

$\frac{3}{2}$

Explanation

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\&\text{two values: one at the point of interest and another succeeding}\&\text{it, and finally that distance between the two. Mathematically,}\&\text{this follows the form:}\&f'(x)=\frac{f(x+h)-f(x)}{h}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{In approximating a derivative, it is generally prudent to minimize}\&\text{the distance between function values. In other words, it’s a}\&\text{good idea to look at two adjacent points. For our problem, that’d}\&\text{be:}\&f(7)\text{ and }f(9)\&\text{Applying the formula above, with }h=2\text{, we find:}\&f'(7)\approx\frac{-18-(-18)}{2}\&f'(7)\approx3\end{align*}$

$\begin{align*}&t&p(t)\&1&32\&6&-26\&11&43\&16&-15\&\text{The table above lists the positions of a particle}\&\text{At various times. Approximate the acceleration at }t=6\&\text{Using a central difference approach.}\end{align*}$

$\frac{127}{25}$

$-\frac{58}{5}$

$\frac{11}{25}$

$\frac{11}{5}$

Explanation

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\&\text{words, you take the derivative of a position function twice}\&\text{to define it. In this problem, we’re not given a function, but}\&\text{ a table of times and positions, requiring an approximation,}\&\text{a 2nd order central difference. This has the form:}\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{This approximation is actually born from two types of first}\&\text{order approximations: the forward and backward differences,}\&\text{which have the forms:}\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\&\text{The difference of these two, divided by our increment, gives}\&\text{a derivative of a derivative.}\&\text{Applying the formula above, with }h=5\text{, we find:}\&a(6)\approx\frac{43+(32)-2(-26)}{5^2}\&a(6)\approx\frac{127}{25}\end{align*}$

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