Calculus II — Integrals - Math

Card 0 of 372

Question

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

Answer

The $(x-1)^{2}$ 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 $\frac{f''(1) }{2} ; (x-1)^{2} = - \frac{1}{2 \ln 5 } (x-1)^{2} = - \frac{1}{ \ln 25 } (x-1)^{2}$

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Question

Find the unit vector of .

Answer

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:

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

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

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

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Question

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).

Answer

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

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

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Question

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).

Answer

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

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

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Question

The above graph depicts a function . Does $\lim_{x\rightarrow 1} f(x)$ exist, and why or why not?

Answer

$\lim_{x\rightarrow 1} f(x)$ exists if and only if $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$ ;

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

As can be seen,

$\lim_{x\rightarrow 1^{+}} f(x) = 1$ and $\lim_{x\rightarrow 1^{-}} f(x) = 1$ ;

therefore, $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$ ,

and $\lim_{x\rightarrow 1} f(x)$ exists.

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Question

Given vector and , solve for .

Answer

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:

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Question

The graph depicts a function . Does $\lim_{x\rightarrow 1} g(x)$ exist?

Answer

$\lim_{x\rightarrow 1}g(x)$ exists if and only if $\lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$ ; the actual value of is irrelevant.

As can be seen, $\lim_{x\rightarrow 1^{+}} g(x) = 1$ and $\lim_{x\rightarrow 1^{-}}g(x) = 1$ ; therefore, $\lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$ , and $\lim_{x\rightarrow 1} g(x)$ exists.

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Question

The above graph depicts a function . Does $\lim_{x\rightarrow 1} h(x)$ exist, and why or why not?

Answer

$\lim_{x\rightarrow 1} h(x)$ exists if and only if $\lim_{x\rightarrow 1^{-}} h(x) = \lim_{x\rightarrow 1^{+}} h(x)$ . As can be seen from the diagram, $\lim_{x\rightarrow 1^{-}} h(x) = 1$ , but $\lim_{x\rightarrow 1^{+}} h(x) = -1$ . Since $\lim_{x\rightarrow 1^{-}} h(x) eq \lim_{x\rightarrow 1^{+}} h(x)$ , $\lim_{x\rightarrow 1} h(x)$ does not exist.

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Question

A function is defined by the following piecewise equation:

$f(x)=\left{\begin{matrix} x^{2}+4x-3, x<3\10x-12, x\geq 3 \end{matrix}\right.$

At , the function is:

Answer

The first step to determine continuity at a point is to determine if the function is defined at that point. When we substitute in 3 for , we get 18 as our -value. is thus defined for this function.

The next step is determine if the limit of the function is defined at that point. This means that the left-hand limit must be equal to the right-hand limit at . Substitution reveals the following:

$\lim_{x\rightarrow 3^{-}}f(x)=(3)^{2}+4(3)-3=9+12-3=18$

$\lim_{x\rightarrow 3^{+}}f(x)=10(3)-12=18$

Both sides of the function, therefore, approach a -value of 18.

Finally, we must ensure that the curve is smooth by checking the limit of the derivative of both sides.

$\lim_{x\rightarrow 3^{-}}f^{'}(x)=2(3)+4=10$

$\lim_{x\rightarrow 3^{+}}f^{'}(x)=10$

Since the function passes all three tests, it is continuous.

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Question

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).

Answer

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

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

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Question

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).

Answer

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

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

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Question

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

Answer

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

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Question

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 $\theta$ ?

Answer

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 $<a_{1}, a_{2}, a_{3}>$ and the components of $\vec{b}$ be $<b_{1}, b_{2}, b_{3}>$ . 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 $<a_{1}, a_{2}, a_{3}>$ . 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

$\left | \vec{a}\right |=\sqrt{(a_{1})^2+(a_{2})^2+(a_3)^2}$

$=\sqrt{(1)^2+(2)^2+(1)^2}=\sqrt{1+4+1}=\sqrt{6}$

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.

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

$\theta=\cos^{-1}(\frac{\sqrt{21}}{14})$ .

The answer is $\cos^{-1}(\frac{\sqrt{21}}{14})$ .

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Question

Find the magnitude of vector :

Answer

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

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

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

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

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

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Question

Given vector and , solve for .

Answer

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

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Question

Given vector and , solve for .

Answer

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

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Question

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

Answer

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:

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

$\sqrt{\left(\frac{5}{13}\right)^2+\left(\frac{-12}{13}\right)^2}$

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

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

$\sqrt{1}$

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

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Question

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

Answer

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

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

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

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

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

$\alpha = \arctan (1)$

$\alpha =45^{\circ}$

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Question

Solve for vector given direction of $45^{\circ}$ and magnitude of .

Answer

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

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

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

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

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Question

Given vector and , solve for .

Answer

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:

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