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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 #262 : Vector

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

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

Possible Answers:

$\vec{v}\cdot\vec{u}=6x^3+4x+4xe^{2x}+2e^{2x}$

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

$\vec{v}\cdot\vec{u}=6x^3+4x+4xe^{2x}+2e^{2x}-sin(x)$

$\vec{v}\cdot\vec{u}=6x^2+4x+4xe^{2x}+2e^{2x}+sin(x)$

Correct answer:

$\vec{v}\cdot\vec{u}=6x^3+4x+4xe^{2x}+2e^{2x}-sin(x)$

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:

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

$\vec{u} =(2x,(4x+2),-1)$

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

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

$\vec{v}\cdot\vec{u}=6x^3+6x+4xe^{2x}+2e^{2x}+(-sin(x)-2x)$

$\vec{v}\cdot\vec{u}=6x^3+4x+4xe^{2x}+2e^{2x}-sin(x)$

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Example Question #602 : Parametric, Polar, And Vector

$\newline\vec{v} =(x^2+2x,5x^3-2x+1,2x+6)\newline\vec{u} =(2x^2+2x+4,x^2+x,x^2+x)\newline$

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

Possible Answers:

$\vec{v}+\vec{u}=(4x^2,5x^3+2x^2,x^2+3x+6)$

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

$\vec{v}+\vec{u}=(3x^2+4x+4,5x^3+x^2-x+1,x^2+3x+6)$

$\vec{v}+\vec{u}=(3x^2+x,5x^3+x^2+5,x^2+2x+6)$

Correct answer:

$\vec{v}+\vec{u}=(3x^2+4x+4,5x^3+x^2-x+1,x^2+3x+6)$

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:

$\vec{v} =(x^2+2x,5x^3-2x+1,2x+6)$

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

$\vec{v}+\vec{u}=(x^2+2x,5x^3-2x+1,2x+6)+(2x^2+2x+4,x^2+x,x^2+x)$

$\vec{v}+\vec{u}=(x^2+2x+2x^2+2x+4,5x^3-2x+1+x^2+x,2x+6+x^2+x)$

$\vec{v}+\vec{u}=(3x^2+4x+4,5x^3+x^2-x+1,x^2+3x+6)$

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Example Question #264 : Vector

$\newline\vec{v} =(5x,16+x,2x^2)\newline\vec{u} =(e^x,2,3x)\newline$

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

Possible Answers:

$\vec{v}\cdot\vec{u}=2x^2+e^x+5x+32$

$\vec{v}\cdot\vec{u}=x^2+2xe^x+4x+32$

$\vec{v}\cdot\vec{u}=2x^2+7x+32$

$\vec{v}\cdot\vec{u}=5xe^x+32+2x+6x^3$

Correct answer:

$\vec{v}\cdot\vec{u}=5xe^x+32+2x+6x^3$

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:

$\vec{v}\cdot\vec{u}=(5x,16+x,2x^2)\cdot(e^x,2,3x)$

$\vec{v}\cdot\vec{u}=5x*e^x+(16+x)*2+2x^2*3x$

$\vec{v}\cdot\vec{u}=5xe^x+32+2x+6x^3$

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Example Question #265 : Vector

$\newline\vec{v} =\left(\left(\frac{\sqrt{2}+1}{2}\right),\frac{2x}{e^{x}},-2\right)\newline\vec{u} =\left(\left(4+x\right),ln(x),2\right)\newline$

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

Possible Answers:

$\vec{v}\cdot\vec{u}=2\sqrt{2}+4+\frac{2xln(x)}{e^x}$

$\vec{v}\cdot\vec{u}=\frac{2xln(x)}{e^x}$

$\vec{v}\cdot\vec{u}=2\sqrt{2}-2+\frac{(\sqrt2+1)}{2}x+\frac{2xln(x)}{e^x}$

$\vec{v}\cdot\vec{u}=2\sqrt{2}+2+\frac{2xln(x)}{e^x}$

Correct answer:

$\vec{v}\cdot\vec{u}=2\sqrt{2}-2+\frac{(\sqrt2+1)}{2}x+\frac{2xln(x)}{e^x}$

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:

$\newline\vec{v} =\left(\left(\frac{\sqrt{2}+1}{2}\right),\frac{2x}{e^{x}},-2\right)\newline\vec{u} =\left(\left(4+x\right),ln(x),2\right)\newline$

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

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

$\vec{v}\cdot\vec{u}=2\sqrt{2}+2 + (\frac{\sqrt{2}+1}{2})*x+\frac{2xln(x)}{e^{x}}+-4$

$\vec{v}\cdot\vec{u}=2\sqrt{2}-2+\frac{(\sqrt2+1)}{2}x+\frac{2xln(x)}{e^x}$

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Example Question #266 : Vector

Find the sum of the vectors and if

$u=\left<3, 7\right>$

$v=\left<7, 10\right>$ .

Possible Answers:

$u+v=\left<17,38\right>$

$u+v=\left<21,70\right>$

$u+v=\left<10,17\right>$

$u+v=\left<4,3\right>$

Correct answer:

$u+v=\left<10,17\right>$

Explanation:

For the two vectors

$a=\left<a_1,a_2\right>$ and $b=\left<b_1,b_2\right>$

The sum of the two vectors a+b is

$\left<a_1, a_2\right> + \left<b_1, b_2\right>= \left<a_1+b_1, a_2+b_2\right>$

For the vectors u and v in this problem the sum u+v is

$\left<3, 7\right> + \left<7, 10\right>$

$= \left<3+7, 7+10\right>$

$=\left <10, 17\right>$

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Example Question #1 : Derivatives

Evaluate the limit using one of the definitions of a derivative.

$\lim_{x \rightarrow -\frac{\pi}{2}} \frac{\sin^2x -1}{x+\frac{\pi}{2}}$

Possible Answers:

$\pi$

Does not exist

$\frac{1}{2}$

Correct answer:

Explanation:

Evaluating the limit directly will produce an indeterminant solution of $\small \tiny{\frac{0}{0}}$ .

The limit definition of a derivative is $\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f'(x)$ . However, the alternative form, $\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=f'(x)$ , better suits the given limit.

Let $f(x)=\sin^2x$ and notice $f\left(-\frac{\pi}{2}\right)=1$ . It follows that ${\lim_{x \rightarrow -\frac{\pi}{2}} \frac{\sin^2-1}{x-\left(-\frac{\pi}{2} \right )} = \frac{d}{dx}\sin^2x}$ .

Thus, the limit is ${2\sin\left(-\frac{\pi}{2} \right ) \cos\left(-\frac{\pi}{2}\right)} = 0.$

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Example Question #2 : Derivatives

Evaluate the limit using one of the definitions of a derivative.

$\lim_{x \rightarrow 1} \frac{x^{49}-1}{x-1}$

Possible Answers:

Does not exist

Correct answer:

Explanation:

Evaluating the derivative directly will produce an indeterminant solution of $\small \tiny{\frac{0}{0}}$ .

Let $f(x)=x^{49}$ and notice f(1)=1 . It follows that ${\lim_{x \rightarrow 1} \frac{x^{49}-1}{x-1} = \frac{d}{dx}x^{49}}$ . Thus, the limit is $49(1)^{48}=49$ .

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Example Question #1 : Definition Of Derivative

Suppose and are differentiable functions, and f(3)=1, f'(3)=4, g(-1)=3, g'(-1)=-1 . Calculate the derivative of h(x) = f(x+2)g(-x) , at x = 1

Possible Answers:

None of the other answers

Correct answer:

None of the other answers

Explanation:

The correct answer is 11.

Taking the derivative of h(x) involves the product rule, and the chain rule.

h'(x)=[f(x+2)][-g'(-x)]+[g(-x)][f'(x+2)]

= -g'(-x)f(x+2)+g(-x)f'(x+2)

Substituting x=1 into both sides of the derivative we get

$h'(1)= -g'(-1)f(3)+g(-1)f'(3) = -1\cdot1 + 3 \cdot 4 = 11$ .

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Example Question #1 : Definition Of Derivative

Evaluate the limit

$\small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}$

without using L'Hopital's rule.

Possible Answers:

$\small -\frac{\sqrt{2}}{2}$

$\small 1$

$\small 2$

$\small \frac{\sqrt{2}}{2}$

Correct answer:

$\small 1$

Explanation:

If we recall the definition of a derivative of a function $\small f(x)$ at a point $\small x$ , one of the definitions is

$\small f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ .

If we compare this definition to the limit

$\small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}$

we see that that this is the limit definition of a derivative, so we need to find the function $\small f(x)$ and the point $\small x$ at which we are evaluating the derivative at. It is easy to see that the function is $\small f(x)=\tan x$ and the point is $\small x=0$ . So finding the limit above is equivalent to finding $\small f'(0)$ .

We know that the derivative is $\small f'(x)=\sec^2(x)$ , so we have

$\small \small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}=f'(0)=\sec^2(0)=1^2=1$ .

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Example Question #4 : Derivatives

Approximate the derivative if $y=\sqrt{x}$ where x=2 .

Possible Answers:

0.35355

0.35346

0.35353

0.35351

0.35359

Correct answer:

0.35355

Explanation:

Write the definition of the limit.

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Substitute x=2 .

$\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0}\frac{\sqrt{2+h}-\sqrt2}{h}$

Since is approaching to zero, it would be best to evaluate when we assume that is progressively decreasing. Let's assume h=0.1, 0.0001, 0.000001 and check the pattern.

$\frac{\sqrt{2+0.1}-\sqrt2}{0.1}=0.349241$

$\frac{\sqrt{2+0.0001}-\sqrt2}{0.0001}=0.353549$

$\frac{\sqrt{2+0.000001}-\sqrt2}{0.000001}=0.353553$

The best answer is: 0.35355

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #262 : Vector

Example Question #602 : Parametric, Polar, And Vector

Example Question #264 : Vector

Example Question #265 : Vector

Example Question #266 : Vector

Example Question #1 : Derivatives

Example Question #2 : Derivatives

Example Question #1 : Definition Of Derivative

Example Question #1 : Definition Of Derivative

Example Question #4 : Derivatives

All Calculus 2 Resources

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