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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : Vector Calculations

Evaluate: $\left \langle3,-1,2 \right \rangle \cdot \left \langle -2,4,5\right \rangle$

Possible Answers:

$\left \langle 1,3,7\right \rangle$

$\left \langle -6,-4,10\right \rangle$

Correct answer:

Explanation:

Let vector $\vec{a}=\left \langle a1,a2,a3\right \rangle$ and $\vec{b}=\left \langle b1,b2,b3\right \rangle$ .

Use the following formula to solve this dot product:

$\vec{a}\cdot\vec{b} = a1\cdot b1+ a2\cdot b2+a3\cdot b3$

Substitute and solve.

$\left \langle3,-1,2 \right \rangle \cdot \left \langle -2,4,5\right \rangle=(3)(-2)+(-1)(4)+(2)(5 )=0$

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Example Question #1 : Vector Calculations

Calculate the cross product: $(i-j)\times-k$

Possible Answers:

i-j

-j+k

i-j-k

i-k

i+j

Correct answer:

i+j

Explanation:

The vectors can be rewritten in the following form:

$\begin{vmatrix} i& j& k\\ 1&-1 &0 \\ 0& 0& -1 \end{vmatrix}$

One method of solving the 3 by 3 determinant is to break this down into 2 by 2 determinants. The determinant of 2 by 2 matrices are computed by ad-bc , such that:

$D=\begin{vmatrix} a&b \\ c& d \end{vmatrix}$

Rewrite the 3 by 3 determinant.

$=i\begin{vmatrix} -1& 0\\ 0&-1 \end{vmatrix}- j\begin{vmatrix} 1&0 \\ 0&-1 \end{vmatrix}+k\begin{bmatrix} 1&-1 \\ 0&0 \end{bmatrix}$

=i[(-1)(-1)-0] -j[(1)(-1)-0]+k[0]

= i[1]-j[-1]

= i+j

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Example Question #1 : Vector Calculations

A sling shoots a rock feet per second at an elevation angle of degrees. What are the horizontal and vertical components in vector form?

Possible Answers:

$\left \langle -9.848,-1.736\right \rangle$

$\left \langle -9.848,1.736\right \rangle$

$\left \langle 9.848,1.736\right \rangle$

$\left \langle -8.391,-5.440\right \rangle$

$\left \langle 8.391,5.440\right \rangle$

Correct answer:

$\left \langle 9.848,1.736\right \rangle$

Explanation:

The horizontal and vertical components are shown below:

$Vx=Vcos\Theta$

$Vy=Vsin\Theta$

Plug in the velocity and the given angle to the equations.

$Vx=Vcos\Theta=10cos(10) = 9.848$

$Vy=Vsin\Theta = 10sin(10)=1.736$

Therefore, the components in vector form is $\left \langle 9.848,1.736\right \rangle$ .

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Example Question #2 : Vector Calculations

Evaluate the dot product of $\left \langle 1,6\right \rangle$ and $\left \langle 1,6\right \rangle$ .

Possible Answers:

$\infty$

Correct answer:

Explanation:

Let vectors $\vec{a}=\left \langle a1,a2\right \rangle$ and $\vec{b}=\left \langle b1,b2\right \rangle$ .

The formula for the dot product is:

$\vec{a}\cdot\vec{b} = a1\cdot b1+ a2\cdot b2$

Follow this formula and simplify.

$\left \langle 1,6\right \rangle\cdot\left \langle 1,6\right \rangle = (1)(1)+(6)(6)=37$

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Example Question #2 : Vector Calculations

Solve: $\left \langle 3,4,5\right \rangle\cdot\left \langle1,2,3 \right \rangle$

Possible Answers:

$\left \langle 2,-4,2\right \rangle$

$\left \langle 3,8,15\right \rangle$

$\left \langle 4,6,8\right \rangle$

Correct answer:

Explanation:

The problem $\left \langle 3,4,5\right \rangle\cdot\left \langle1,2,3 \right \rangle$ is in the form of a dot product. The final answer must be an integer, and not in vector form.

Write the formula for the dot product.

$\left \langle a_1,a_2,a_3\right \rangle\cdot\left \langle b_1,b_2,b_3\right \rangle=a_1b_1+a_2b_2+a_3b_3$

Substitute the givens and solve.

(3)(1)+(4)(2)+(5)(3)=3+8+15=26

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Example Question #8 : Vector Calculations

Suppose $\vec{a}= \left \langle 2,4,-3\right \rangle$ . Find the magnitude of $2\vec{a}$ .

Possible Answers:

$\sqrt{69}$

$2\sqrt{29}$

$\sqrt{29}$

Correct answer:

$2\sqrt{29}$

Explanation:

Calculate $2\vec{a}$ .

$\vec{a}= \left \langle 2,4,-3\right \rangle$

$2\vec{a}= \left \langle 4,8,-6\right \rangle$

Find the magnitude.

$|2\vec{a}|=\sqrt{4^2+8^2+(-6)^2}=\sqrt{16+64+36}=\sqrt{116}=2\sqrt{29}$

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Example Question #161 : Vector

Two particles move freely in two dimensional space. The first particle's location as a function of time is $\small \left<3t^{2}-t,4\right>$ , and the second particle's location is $\small \small \left<6,t+1/t\right>$ . Will the particles ever collide for $\small t>0$ ?

Possible Answers:

Impossible to determine

No, because the particles' $\small x$ and $\small y$ coordinates are never the same simultaneously at any instant in time.

Yes, because the particles have the same $\small x$ or $\small y$ component (not necessarily simultaneously).

No, because the particles never have the same $\small x$ or $\small y$ component (not necessarily simultaneously).

Yes, because the particles' $\small x$ and $\small y$ coordinates are the same simultaneously at a certain instant in time.

Correct answer:

No, because the particles' $\small x$ and $\small y$ coordinates are never the same simultaneously at any instant in time.

Explanation:

In order for the particles to collide, their $\small x$ and $\small y$ coordinates must be equal simultaneously. In order to check if this happens, we can set the particles' $\small x$ -coordinates and $\small y$ -coordinates equal to each other.

Let's start with the $\small x$ -coordinate:

$\small \small \small 3t^{2}-t=6\rightarrow 3t^2-t-6=0$ .

Using the quadratic formula

$\small \small t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ , we get $\small \small t=\frac{1+\sqrt{73}}{6}\approx1.59$

the other root is negative, so it can be discarded since $\small t>0$ .

Now let's do the $\small y$ -coordinate: $\small \small t+1/t=4\rightarrow t^2+1=4t\rightarrow t^2-4t+1=0$ . Use the quadratic formula again to solve for $\small t$ , and you'll get $\small 2\pm\sqrt{3}$ (these roots are approximately $\small 0.27$ and $\small 3.73$ ). The particles never have the same $\small x$ and $\small y$ coordinate simultaneously, so they do not collide.

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Example Question #1 : Vector Calculations

$\vec{a}=<3,2,1>$

$\vec{b}=<1,0,10>$

Calculate $\vec{a}\cdot\vec{b}$

Possible Answers:

Correct answer:

Explanation:

$\vec{a}=<3,2,1>$

$\vec{b}=<1,0,10>$

$\vec{a}\cdot\vec{b}$ is simply the dot product of these two vectors. Mathematically, this is calculated as follows.

$(3\cdot 1)+(2\cdot 0)+(1\cdot 10)$ =13

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Example Question #11 : Vector Calculations

Find the dot product of $a=\left \langle 3,4,7\right \rangle$ and $b=\left \langle -1,5,6\right \rangle$ .

Possible Answers:

Correct answer:

Explanation:

To find the dot product of $a=\left \langle 3,4,7\right \rangle$ and $b=\left \langle -1,5,6\right \rangle$ , calculate the sum of the products of the vectors' corresponding components:

$\left \langle 3,4,7\right \rangle\times \left \langle -1,5,6\right \rangle$

$=(3\times -1)+ (4\times 5)+(7\times 6)$

=-3+ 20+42

=-3+ 62

=59

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Example Question #11 : Vector Calculations

Find the dot product of $a=\left \langle 5,-5,2\right \rangle$ and $b=\left \langle 2,3,8\right \rangle$ .

Possible Answers:

Correct answer:

Explanation:

To find the dot product of $a=\left \langle 5,-5,2\right \rangle$ and $b=\left \langle 2,3,8\right \rangle$ , calculate the sum of the products of the vectors' corresponding components:

$\left \langle 5,-5,2\right \rangle\times \left \langle 2,3,8\right \rangle$

$=(5\times 2)+ (-5\times 3)+(2\times 8)$

=10+ (-15)+16

=10+ 1

=11

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #3 : Vector Calculations

Example Question #1 : Vector Calculations

Example Question #1 : Vector Calculations

Example Question #2 : Vector Calculations

Example Question #2 : Vector Calculations

Example Question #8 : Vector Calculations

Example Question #161 : Vector

Example Question #1 : Vector Calculations

Example Question #11 : Vector Calculations

Example Question #11 : Vector Calculations

All Calculus 2 Resources

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