Calculus II — Integrals - Math

Card 0 of 372

Give the term of the Taylor series expansion of the function $f (x) = \log_5 x$ about .

Tap to see back →

The term of a Taylor series expansion about is

$\frac{f ' ' (1)}{2!}$ $(x-1)^{2}$= $\frac{f ' ' $(1)}{2}$(x-1)^{2}$ .

We can find by differentiating twice in succession:

$f (x) = \log_5 x$

$f'(x) = $\frac{1}{x \ln 5}$$

$f'(x) =\frac{1}{ \ln 5}$ \cdot $x^{-1}$$

$f''(x) = $\frac{\mathrm{d}$ }{\mathrm{d} x}\left($\frac{1}{ \ln 5}$ \cdot $x^{-1}$ \right)$

$f''(x) =\frac{1}{ \ln 5}$\cdot $\frac{\mathrm{d}$ }{\mathrm{d} x}\left ( $x^{-1}$ \right )$

$f''(x) =\frac{1}{ \ln 5}$ \cdot \left( -1 \cdot $x^{-2}$ \right)$

$f''(x) =- $\frac{1}{ \ln 5 \cdot $x^{2}$$}$

$f''(1) =- $\frac{1}{ \ln 5 \cdot $1^{2}$$} = - $\frac{1}{ \ln 5 }$$

so the term is $\frac{f''(1) }{2}$ ; $(x-1)^{2}$ = - $\frac{1}{2 \ln 5 }$ $(x-1)^{2}$ = - $\frac{1}{ \ln 25 }$ $(x-1)^{2}$

Find the unit vector of .

Tap to see back →

To solve for the unit vector, the following formula must be used:

$\frac{v}{||v||}$

$\left | v \right $|=$\sqrt{(5)^2$$+(-12)^2$}$$

$\left | v \right |=$\sqrt{25+144}$$

$\left | v \right |=$\sqrt{169}$$

$\left | v \right |=13$

unit vector:

$\frac{<5,-12>}{13}$

$=\frac{1}{13}$*<5,-12>$

$=<$\frac{5}{13}$,$\frac{-12}{13}$>$

The polar coordinates of a point are $\left(2.1, $\frac{7\pi }{3}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Tap to see back →

Given the polar coordinates $(r,\theta )$ , the -coordinate is $y= r \sin \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = $\frac{7\pi }{3}$$ :

$y = r \sin \theta = 2.1 \cdot \sin $\frac{7\pi }{3}$ \approx 2.1 \cdot 0.8660 \approx 1.82$

The polar coordinates of a point are $\left(2.1, $\frac{7\pi }{3}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Tap to see back →

Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = $\frac{7\pi }{3}$$ :

$x = r \cos \theta = 2.1 \cdot \cos $\frac{7\pi }{3}$ = 2.1 \cdot 0.5 = 1.05$

The above graph depicts a function . Does exist, and why or why not?

Tap to see back →

exists if and only if ;

the actual value of is irrelevant, as is whether is continuous there.

As can be seen,

and ;

therefore, ,

and exists.

Given vector and , solve for .

Tap to see back →

To solve for , We need to first multiply into vector to find and multiply into vector to find ; then we need to subtract the components in the vector and the components together:

The graph depicts a function . Does exist?

Tap to see back →

exists if and only if ; the actual value of is irrelevant.

As can be seen, and ; therefore, , and exists.

The above graph depicts a function . Does exist, and why or why not?

Tap to see back →

exists if and only if . As can be seen from the diagram, , but . Since , does not exist.

The polar coordinates of a point are $\left(1.2, $\frac{2\pi }{5}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Tap to see back →

Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 1.2, \theta = $\frac{2\pi }{5}$$ :

$x = r \cos \theta = 1.2 \cdot \cos $\frac{2\pi }{5}$ \approx 1.2 \cdot 0.3090 \approx 0.37$

The polar coordinates of a point are $\left(1.2, $\frac{2\pi }{5}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Tap to see back →

Given the polar coordinates $(r,\theta )$ , the -coordinate is $y= r \sin \theta$ . We can find this coordinate by substituting $r = 1.2, \theta = $\frac{2\pi }{5}$$ :

$y = r \cos \theta = 1.2 \cdot \sin $\frac{2\pi }{5}$ \approx 1.2 \cdot 0.9511 \approx 1.14$

Let $\vec{a}, \vec{b},\vec{c}$ be vectors. All of the following are defined EXCEPT:

Tap to see back →

The cross product of two vectors (represented by "x") requires two vectors and results in another vector. By contrast, the dot product (represented by " $\cdot$ ") between two vectors requires two vectors and results in a scalar, not a vector.

If we were to evaluate $(\vec{a}\cdot \vec{b})\times\vec{c}$ , we would first have to evaluate $\vec{a}\cdot\vec{b}$ , which would result in a scalar, because it is a dot product.

However, once we have a scalar value, we cannot calculate a cross product with another vector, because a cross product requires two vectors. For example, we cannot find the cross product between 4 and the vector <1, 2, 3>; the cross product is only defined for two vectors, not scalars.

The answer is $(\vec{a}\cdot \vec{b})\times\vec{c}$ .

Let $\vec{}$ $\vec{a}$ and $\vec{b}$ be the following vectors: $\vec{a}=\left \langle 1,2,1\right \rangle$ and $\vec{b}=\left \langle 2,3,-1\right \rangle$ . If $\theta$ is the acute angle between the vectors, then which of the following is equal to $\theta$ ?

Tap to see back →

The cosine of the acute angle between two vectors is given by the following formula:

$\cos\theta=\frac{\vec{a}$\cdot \vec{b}}{\left | \vec{a}\right | \left | \vec{b} \right |}$ , where $\vec{a} \cdot \vec{b}$ represents the dot product of the two vectors, $\left | \vec{a} \right |$ is the magnitude of vector a, and $\left | \vec{b} \right |$ is the magnitude of vector b.

First, we will need to compute the dot product of the two vectors. Let's say we have two general vectors in space (three dimensions), $\vec{a}$ and $\vec{b}$ . Let the components of $\vec{a}$ be and the components of $\vec{b}$ be . Then the dot product $\vec{a} \cdot \vec{b}$ is defined as follows:

$\vec{a} \cdot $\vec{b}=a_{1}$$(b_{1}$$)+a_{2}$$(b_{2}$$)+a_{3}$$(b_{3}$)$ .

Going back to the original problem, we can use this definition to find the dot product of $\vec{a}=<1,2,1>$ and $\vec{b}=<-2,3,-1>$ .

$\vec{a} \cdot $\vec{b}=a_{1}$$(b_{1}$$)+a_{2}$$(b_{2}$$)+a_{3}$$(b_{3}$)$

The next two things we will need to compute are $\left | \vec{a} \right |$ and $\left | \vec{b} \right |$ .

Let the components of a general vector $\vec{a}$ be . Then $\left | \vec{a} \right |$ is defined as $\left | \vec{a}\right |=$\sqrt{(a_${1}$)^2$+(a_${2})^2$+(a_$3)^2$}$ .

Thus, if $\vec{a}=<1,2,1>$ and $\vec{b}=<-2,3,-1>$ , then

$\left | \vec{a}\right |=$\sqrt{(a_${1}$)^2$+(a_${2})^2$+(a_$3)^2$}$

and $\left | \vec{b}\right |=$\sqrt{(b_${1}$)^2$+(b_${2})^2$+(b_$3)^2$$}=$\sqrt{(-2)^2$$+(3)^2$$+(-1)^2$}$$

$=$\sqrt{4+9+1}$=$\sqrt{14}$$ .

Now, we put all of this information together to find the cosine of the angle between the two vectors.

$\cos\theta=\frac{\vec{a}$\cdot \vec{b}}{\left | \vec{a}\right | \left | \vec{b} \right |}=\frac{3}{\sqrt{6}$\cdot$\sqrt{14}$}$

We just need to simplify this.

$$\frac{3}{\sqrt{6}$\cdot$\sqrt{14}$}=\frac{3}{\sqrt{6\cdot14}$}=\frac{3}{\sqrt{2\cdot3\cdot2\cdot7}$}$

$=\frac{3}{2\sqrt{21}$}$ .

In order to get it completely simplified, we have to rationalize the denominator by multiplying the numerator and denominator by the sqare root of 21.

$$\frac{3}{2\sqrt{21}$}=\frac{3\cdot\sqrt{21}$}{2$\sqrt{21}$\cdot$\sqrt{21}$}=\frac{3\sqrt{21}$}{2(21)}$

$=\frac{\sqrt{21}$}{2\cdot7}=\frac{\sqrt{21}$}{14}$ .

We just have one more step. We need to solve for the value of the angle. In order to do this, we can take the inverse cosine of both sides of the equation.

$\cos\theta=\frac{\sqrt{21}$}{14}$

.

The answer is .

Find the magnitude of vector :

Tap to see back →

To solve for the magnitude of a vector, we use the following formula:

$\left | v \right $|=$\sqrt{x^2$$+y^2$}$$

$\left | v \right $|=$\sqrt{(3)^2$$+(-2)^2$}$$

$\left | v \right |=$\sqrt{9+}$4$

$\left | v \right |=$\sqrt{13}$$

Given vector and , solve for .

Tap to see back →

To solve for , we need to add the components in the vector and the components together:

Given vector and , solve for .

Tap to see back →

To solve for , we need to subtract the components in the vector and the components together:

Is $\langle $\frac{5}{13}$,$\frac{-12}{13}$\rangle$ a unit vector?

Tap to see back →

To verify where a vector is a unit vector, we must solve for its magnitude. If the magnitude is equal to 1 then the vector is a unit vector:

$<$\frac{5}{13}$,$\frac{-12}{13}$>$

$$\sqrt{\left(\frac{25}{169}\right)+\left($\frac{144}{169}\right)$

$$\sqrt{\frac{169}{169}$}$

$\sqrt{1}$

$<$\frac{5}{13}$,$\frac{-12}{13}$>$ is a unit vector because magnitude is equal to .

Given vector . Solve for the direction (angle) of the vector:

Tap to see back →

To solve for the direction of a vector, we use the following formula:

$tan\alpha$ = $\frac{b}{a}$

with the vector being $u=\left \langle a,b \right \rangle$

$u=\left \langle 3,3 \right \rangle$

$\tan\alpha=\frac{3}{3}$=1$

$\alpha = \arctan (1)$

$\alpha $=45^{\circ}$$

Solve for vector given direction of and magnitude of .

Tap to see back →

To solve for a vector with the magnitude and direction given, we use the following formula:

$v=(8\cos $45^{\circ}$)i $+(8\sin45^{\circ}$)j$

$v=4$\sqrt{2}$i+4$\sqrt{2}$j$

$v=\left \langle 4$\sqrt{2}$,4$\sqrt{2}$} \right \rangle$

Given vector and , solve for .

Tap to see back →

To solve for , We need to multiply into vector to find ; then we need to subtract the components in the vector and the components together:

Find the magnitude of .

Tap to see back →

therefore the vector is

To solve for the magnitude:

$\sqrt{(36)+(36)}$

$\sqrt{72}$

$6$\sqrt{2}$$