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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 #591 : Parametric, Polar, And Vector

What is the cross product of $a=\left \langle 5,2,8\right \rangle$ and $b=\left \langle -2,-4,5\right \rangle$ ?

Possible Answers:

$\left \langle 42, -41,-16 \right \rangle$

$\left \langle 42, 41,-16 \right \rangle$

$\left \langle 42, -41,16 \right \rangle$

$\left \langle -42, 41,16 \right \rangle$

$\left \langle -42, -41,16 \right \rangle$

Correct answer:

$\left \langle 42, 41,-16 \right \rangle$

Explanation:

In order to find the cross product of two three-dimensional vectors, we must find the determinant of a matrix comprised of the vectors' elements. That is, if $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_{1},b_{2},b_{3}\right \rangle$ , then

$a\times b=\begin{Bmatrix} i&j &k \\ a_{1}&a_{2} &a_{3} \\ b_{1}&b_{2} &b_{3} \end{Bmatrix}=\left \langle a_{2}b_{3}-a_{3}b_{2}, a_{1}b_{3}-a_{3}b_{1},a_{1}b_{2}-a_{2}b_{1} \right \rangle$ .

Given $a=\left \langle 5,2,8\right \rangle$ and $b=\left \langle -2,-4,5\right \rangle$ , the cross product $a \times b$ is:

$\begin{Bmatrix} i& j&k \\ 5&2 &8 \\ -2&-4 &5 \end{Bmatrix}$

$=\left \langle (2)(5)-(8)(-4), (5)(5)-(8)(-2),(5)(-4)-(2)(-2) \right \rangle$

$=\left \langle 10+32, 25+16,-20+4 \right \rangle$

$=\left \langle 42, 41,-16 \right \rangle$

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Example Question #592 : Parametric, Polar, And Vector

What is the cross product of $a=\left \langle 1,-1,2\right \rangle$ and $b=\left \langle 2,1,-3\right \rangle$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

$a\times b=\begin{Bmatrix} i&j &k \\ a_{1}&a_{2} &a_{3} \\ b_{1}&b_{2} &b_{3} \end{Bmatrix}=\left \langle a_{2}b_{3}-a_{3}b_{2}, a_{1}b_{3}-a_{3}b_{1},a_{1}b_{2}-a_{2}b_{1} \right \rangle$ .

Given $a=\left \langle 1,-1,2\right \rangle$ and $b=\left \langle 2,1,-3\right \rangle$ ?the cross product $a \times b$ is:

$\begin{Bmatrix} i& j&k \\ 1&-1 &2 \\ 2&1 &-3 \end{Bmatrix}$

$=\left \langle (-1)(-3)-(2)(1),(1)(-3)-(2)(2),(1)(1)-(-1)(2) \right \rangle$

$=\left \langle 3-2,-3-4,1+2 \right \rangle$

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

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Example Question #1111 : Calculus Ii

Vector $\vec{v} = (5,11,-3)$ .

Calculate the magnitude, or $\left \| \vec{v} \right \|$ , of $\vec{v}$

Possible Answers:

$\left \| \vec{v} \right \|=\sqrt{156}$

$\left \| \vec{v} \right \|=\sqrt{154}$

$\left \| \vec{v} \right \|=\sqrt{155}$

$\left \| \vec{v} \right \|=\sqrt{157}$

Correct answer:

$\left \| \vec{v} \right \|=\sqrt{155}$

Explanation:

Calculating magnitude:

If $\vec{v} = (v_1,v_2,v_3)$

Then the magnitude of $\vec{v}$ is $\sqrt{v_1^2 + v_2^2 + v_3^2}$

Note: Magnitude, length, and norm are synonymous.

$\vec{v} = (5,11,-3)$

$\left \| \vec{v} \right \|=\sqrt{5^2+11^2+(-3)^2}$

$\left \| \vec{v} \right \|=\sqrt{25+121+9}$

$\left \| \vec{v} \right \|=\sqrt{155}$

$\left \| \vec{v} \right \|$ cannot be further reduced, so the magnitude of $\vec{v}$

is $\sqrt{155}$

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Example Question #251 : Vector

Vector $\vec{v} = (-6,6,9)$ .

Calculate the magnitude, or $\left \| \vec{v} \right \|$ , of $\vec{v}$

Possible Answers:

$\left \| \vec{v} \right \|=\sqrt{150}$

$\left \| \vec{v} \right \|=\sqrt{151}$

$\left \| \vec{v} \right \|=9$

$\left \| \vec{v} \right \|=\sqrt{153}$

Correct answer:

$\left \| \vec{v} \right \|=\sqrt{153}$

Explanation:

Calculating magnitude:

If $\vec{v} = (v_1,v_2,v_3)$

Then the magnitude of $\vec{v}$ is $\sqrt{v_1^2 + v_2^2 + v_3^2}$

Note: Magnitude, length, and norm are synonymous.

$\vec{v} = (-6,6,9)$

$\left \| \vec{v} \right \|=\sqrt{(-6)^2+6^2+9^2}$

$\left \| \vec{v} \right \|=\sqrt{36+36+81}$

$\left \| \vec{v} \right \|=\sqrt{153}$

$\left \| \vec{v} \right \|$ cannot be further reduced, and so $\left \| \vec{v} \right \|=\sqrt{153}$

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Example Question #1112 : Calculus Ii

Vector $\vec{v} = (sin(x),cos(x),-1)$ .

Calculate the magnitude of $\vec{v}$ , as in $\left \| \vec{v} \right \|$

Possible Answers:

$\left \| \vec{v} \right \|=0$

$\left \| \vec{v} \right \|=1$

$\left \| \vec{v} \right \|=\sqrt{2}$

$\left \| \vec{v} \right \|=2$

Correct answer:

$\left \| \vec{v} \right \|=\sqrt{2}$

Explanation:

Calculating magnitude:

If $\vec{v} = (v_1,v_2,v_3)$

Then the magnitude of $\vec{v}$ is $\sqrt{v_1^2 + v_2^2 + v_3^2}$

Note: Magnitude, length, and norm are synonymous.

Note: sin(x)^2 + cos(x)^2 = 1 , and this is a basic trig identity that you should know.

$\vec{v} = (sin(x),cos(x),-1)$

$\left \| \vec{v} \right \|=\sqrt{sin(x)^2+cos(x)^2+(-1)^2}$

$\left \| \vec{v} \right \|=\sqrt{sin(x)^2+cos(x)^2+1}$

$\left \| \vec{v} \right \|=\sqrt{1 + 1}$

$\left \| \vec{v} \right \|=\sqrt{2}$

$\left \| \vec{v} \right \|$ cannot be further reduced, and so $\left \| \vec{v} \right \|=\sqrt{2}$

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Example Question #252 : Vector

Vector $\vec{v} = (2x^3,3x^{3},\sqrt{12}x^3)$ .

Calculate the magnitude, or $\left \| \vec{v} \right \|$ , of $\vec{v}$

Possible Answers:

$\left \| \vec{v} \right \|=6x^3$

$\left \| \vec{v} \right \|=7x^3$

$\left \| \vec{v} \right \|=5x^3$

$\left \| \vec{v} \right \|=4x^3$

Correct answer:

$\left \| \vec{v} \right \|=5x^3$

Explanation:

Calculating magnitude:

If $\vec{v} = (v_1,v_2,v_3)$

Then the magnitude of $\vec{v}$ is $\sqrt{v_1^2 + v_2^2 + v_3^2}$

Note: Magnitude, length, and norm are synonymous.

$\vec{v} = (2x^3,3x^{3},\sqrt{12}x^3)$

$\left \| \vec{v} \right \|=\sqrt{(2x^3)^2+(3x^3)^2+(\sqrt{12}x^3)^2}$

$\left \| \vec{v} \right \|=\sqrt{4x^6+9x^6+12x^6}$

$\left \| \vec{v} \right \|=\sqrt{25x^6}$

$\left \| \vec{v} \right \|=5x^3$

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Example Question #1 : Vector Form

$\newline\vec{v} =(x^2,-x,5)\newline\vec{u} =(2,3x,11)\newline$

Calculate $\vec{v} + \vec{u}$

Possible Answers:

$\vec{v}+\vec{u}=(x^2 + 2,2x,16)$

$\vec{v}+\vec{u}=(x^2,2x,16)$

$\vec{v}+\vec{u}=(x^2 + 2,0,16)$

$\vec{v}+\vec{u}=(x^2,2x,3x)$

Correct answer:

$\vec{v}+\vec{u}=(x^2 + 2,2x,16)$

Explanation:

Calculate the sum of vectors.

In general,

$\vec{v} = (v_1,v_2,v_3)$

$\vec{u} = (u_1,u_2,u_3)$

$\vec{v} + \vec{u} = (v_1 + u_1, v_2 + u_2,v_3 + u_3)$

Solution:

$\newline\vec{v} =(x^2,-x,5)\newline\vec{u} =(2,3x,11)\newline$

$\vec{v}+\vec{u}=(x^2,-x,5)+(2,3x,11)$

$\vec{v}+\vec{u}=(x^2+2,-x+3x,5+11)$

$\vec{v}+\vec{u}=(x^2 + 2,2x,16)$

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Example Question #253 : Vector

$\newline\vec{v} =(2x+3,3x^2+1,e^x+2)\newline\vec{u} =(-2x+4,-11x^2+2,7e^x+x^2+3)\newline$

Calculate $\vec{v} + \vec{u}$

Possible Answers:

$\vec{v}+\vec{u}=(7,-8x^2+3,8e^x+x^2+5)$

$\vec{v}+\vec{u}=(7,-8x^2,8e^x+x^2+5)$

$\vec{v}+\vec{u}=(7,-8x^2+3,e^{8x}+x^2+5)$

$\vec{v}+\vec{u}=(7,-8x+3,8e^x+x^2+4)$

Correct answer:

$\vec{v}+\vec{u}=(7,-8x^2+3,8e^x+x^2+5)$

Explanation:

Calculate the sum of vectors.

In general,

$\vec{v} = (v_1,v_2,v_3)$

$\vec{u} = (u_1,u_2,u_3)$

$\vec{v} + \vec{u} = (v_1 + u_1, v_2 + u_2,v_3 + u_3)$

Solution:

$\vec{v} =(2x+3,3x^2+1,e^x+2)$

$\vec{u} =(-2x+4,-11x^2+2,7e^x+x^2+3)$

$\vec{v}+\vec{u}=(2x+3,3x^2+1,e^x+2)+(-2x+4,-11x^2+2,7e^x+x^2+3)$

$\vec{v}+\vec{u}=(2x+3+-2x+4,3x^2+1+-11x^2+2,e^x+2+7e^x+x^2+3)$

$\vec{v}+\vec{u}=(7,-8x^2+3,8e^x+x^2+5)$

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Example Question #254 : Vector

$\newline\vec{v} =(sin(x)+2x,2x^2+3x-3,e^{2x}+3)\newline\vec{u} =(-2x+3,2x^4+4x,e^{3x}+2)\newline$

Calculate $\vec{v} + \vec{u}$

Possible Answers:

$\vec{v}+\vec{u}=(sin(x)+3,2x^4+2x^2+7x-3,5e^{x}+5)$

$\vec{v}+\vec{u}=(sin(x)+3,2x^4+2x^2+7x-3,e^{2x}+e^{3x}+5)$

$\vec{v}+\vec{u}=(sin(x)+3,2x^4+2x^2+7x-2,e^{2x}+e^{3x}+4)$

$\vec{v}+\vec{u}=(sin(x)+3,2x^4+2x^2+7x-3,e^{2x}+e^{3x})$

Correct answer:

$\vec{v}+\vec{u}=(sin(x)+3,2x^4+2x^2+7x-3,e^{2x}+e^{3x}+5)$

Explanation:

Calculate the sum of vectors.

In general,

$\vec{v} = (v_1,v_2,v_3)$

$\vec{u} = (u_1,u_2,u_3)$

$\vec{v} + \vec{u} = (v_1 + u_1, v_2 + u_2,v_3 + u_3)$

Solution:

$\vec{v} =(sin(x)+2x,2x^2+3x-3,e^{2x}+3)$

$\vec{u} =(-2x+3,2x^4+4x,e^{3x}+2)$

$\vec{v}+\vec{u}=(sin(x)+2x,2x^2+3x-3,e^{2x}+3)+(-2x+3,2x^4+4x,e^{3x}+2)$

$\vec{v}+\vec{u}=(sin(x)+2x+-2x+3,2x^2+3x-3+2x^4+4x,e^{2x}+3+e^{3x}+2)$

$\vec{v}+\vec{u}=(sin(x)+3,2x^4+2x^2+7x-3,e^{2x}+e^{3x}+5)$

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Example Question #601 : Parametric, Polar, And Vector

$\newline\vec{v} =(5ln(x),2+2x-3x^2,ln(x)+1)\newline\vec{u} =(4+2x,3,\frac{1}{ln(x)+1})\newline$

Calculate $\vec{v} \cdot \vec{u}$

Possible Answers:

$\vec{v}\cdot\vec{u}=10ln(x)+10xln(x)-9x^2+7$

$\vec{v}\cdot\vec{u}=20ln(x)+10xln(x)-9x^2+6x+6$

$\vec{v}\cdot\vec{u}=20ln(x)+ln(10x)-9x^2+6x+6$

$\vec{v}\cdot\vec{u}=20ln(x)+10xln(x)-9x^2+6x+7$

Correct answer:

$\vec{v}\cdot\vec{u}=20ln(x)+10xln(x)-9x^2+6x+7$

Explanation:

Calculate the dot product of the 2 vectors.

In general,

$\vec{v} = (v_1,v_2,v_3)$

$\vec{u} = (u_1,u_2,u_3)$

$\vec{v} \cdot \vec{u} = v_1 * u_1 + v_2 * u_2 + v_3 * u_3$

Solution:

$\vec{v} =(5ln(x),2+2x-3x^2,ln(x)+1)$

$\vec{u} =(4+2x,3,\frac{1}{ln(x)+1})$

$\vec{v}\cdot\vec{u}=(5ln(x),2+2x-3x^2,ln(x)+1)\cdot(4+2x,3,\frac{1}{ln(x)+1})$

$\vec{v}\cdot\vec{u}=5ln(x)*4+2x+2+2x-3x^2*3+(ln(x)+1)*\frac{1}{ln(x)+1}$

$\vec{v}\cdot\vec{u}=20ln(x)+10xln(x)+6+6x-9x^2+1$

$\vec{v}\cdot\vec{u}=20ln(x)+10xln(x)-9x^2+6x+7$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #591 : Parametric, Polar, And Vector

Example Question #592 : Parametric, Polar, And Vector

Example Question #1111 : Calculus Ii

Example Question #251 : Vector

Example Question #1112 : Calculus Ii

Example Question #252 : Vector

Example Question #1 : Vector Form

Example Question #253 : Vector

Example Question #254 : Vector

Example Question #601 : Parametric, Polar, And Vector

All Calculus 2 Resources

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