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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 #16 : Limits And Continuity

Screen shot 2015 08 19 at 5.23.24 pm

Given the above graph of f(x) , over which of the following intervals is f(x) continuous?

Possible Answers:

$(-\infty,0)\cup(0,\infty)$

$(-\infty,-1)\cup(-1,\infty)$

$(-\infty,2)\cup(2,\infty)$

$(-\infty,1)\cup(1,\infty)$

$(-\infty,-2)\cup(-2,0)\cup(0,\infty)$

Correct answer:

$(-\infty,0)\cup(0,\infty)$

Explanation:

For a function f(x) to be continuous at a given point (a, f(a)) , it must meet the following two conditions:

1.) The point (a, f(a)) must exist, and

2.) $\lim_{x\rightarrow a}f(x)=f(a)$ .

Examining the above graph, f(x) is continuous at every possible value of except for , x=0 . Thus, f(x) is continuous on the interval $(-\infty,0)\cup(0,\infty)$ .

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Example Question #511 : Limits

Screen shot 2015 08 19 at 5.43.00 pm

Given the above graph of f(x) , over which of the following intervals is f(x) continuous?

Possible Answers:

$(-\infty,-4)\cup(-4,\infty)$

$(-\infty,-1)\cup(-1,\infty)$

$(-\infty,-3)\cup(-3,\infty)$

$(-\infty,0)\cup(0,\infty)$

$(-\infty,-2)\cup(-2,\infty)$

Correct answer:

$(-\infty,-3)\cup(-3,\infty)$

Explanation:

For a function f(x) to be continuous at a given point (a, f(a)) , it must meet the following two conditions:

1.) The point (a, f(a)) must exist, and

2.) $\lim_{x\rightarrow a}f(x)=f(a)$ .

Examining the above graph, f(x) is continuous at every possible value of except for , x=-3 . Thus, f(x) is continuous on the interval $(-\infty,-3)\cup(-3,\infty)$ .

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Example Question #511 : Calculus Ii

Screen shot 2015 08 21 at 11.18.01 am

Given the above graph of f(x) , over which of the following intervals is f(x) continuous?

Possible Answers:

$(-\infty,0)$

$(0,\infty)$

$(-\infty,0)\cup(0,\infty)$

$(1,\infty)$

$(-\infty,1)\cup(1,\infty)$

Correct answer:

$(0,\infty)$

Explanation:

For a function f(x) to be continuous at a given point (a, f(a)) , it must meet the following two conditions:

1.) The point (a, f(a)) must exist, and

2.) $\lim_{x\rightarrow a}f(x)=f(a)$ .

Examining the above graph, f(x) is continuous at every possible value of except for $x\leq0$ . Thus, f(x) is continuous on the interval $(0,\infty)$ .

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Example Question #1 : Parametric, Polar, And Vector Functions

Rewrite as a Cartesian equation:

$x = t^{2} + 2t + 1, y = t^{2} - 2t + 1, t \in [-1, 1]$

Possible Answers:

y = x - 2

y = x+2

$y = x + 4 - 4\sqrt{x}$

$y = x + 4 + 4\sqrt{x}$

$y = x - 4 - 4\sqrt{x}$

Correct answer:

$y = x + 4 - 4\sqrt{x}$

Explanation:

$x = t^{2} + 2t + 1$

$x = \left ( t + 1\right )^{2}$

$\pm \sqrt{x} = t + 1$

$\sqrt{x} = t + 1$ or $-\sqrt{x} = t + 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so t + 1 is nonnegative; we choose

$\sqrt{x} = t + 1$ .

Also,

$y = t^{2} - 2t + 1$

$y= \left ( t - 1\right )^{2}$

$\sqrt{y} = \pm \left ( t - 1\right )$

$\sqrt{y} = t- 1$ or $-\sqrt{y} = t- 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so t - 1 is nonpositive; we choose

$-\sqrt{y} = t- 1$

or equivalently,

$\sqrt{y} = -t+ 1$

to make t - 1 nonpositive.

Then,

$\sqrt{x}+ \sqrt{y} = (t + 1) + (-t + 1)$

and

$\sqrt{x}+ \sqrt{y} = 2$

$\sqrt{y} = 2 - \sqrt{x}$

$y =\left ( 2 - \sqrt{x} \right )^2$

$y = 4 - 4\sqrt{x} + \left (\sqrt{x} \right )^{2}$

$y = x + 4 - 4\sqrt{x}$

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Example Question #1 : Parametric

Write in Cartesian form:

$x = \cos 2t, y = \sin t$

Possible Answers:

x + 2y = 1

x - 2y = 1

$x - 2y^{2} = 1$

$x + 2y^{2} = 1$

$x = 2y^{2}$

Correct answer:

$x + 2y^{2} = 1$

Explanation:

Rewrite using the double-angle formula:

$x = \cos 2t =1 - 2\sin ^{2} t = 1 - 2y^{2}$

Then

$x = 1 - 2y^{2}$

$x + 2y^{2} = 1$

which is the correct choice.

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Example Question #2 : Parametric

Write in Cartesian form:

$x = 3t, y = 2t^{2}$

Possible Answers:

$\frac{x^{2}}{9} - \frac{y^{2}}{2} = 1$

$\frac{x^{2}}{2} - \frac{y^{2}}{9} = 1$

$x^{2}+ y^{2} = 36$

$y = \frac{2}{9} x^{2}$

$y = \frac{9}{2} x^{2}$

Correct answer:

$y = \frac{2}{9} x^{2}$

Explanation:

x = 3t , so

$x^{2} = (3t)^{2} = 9t^{2} = \frac{9}{2} \cdot 2t^{2} = \frac{9}{2} y$ .

$\frac{9}{2} y= x^{2}$ , so

$\frac{2}{9}\cdot \frac{9}{2} y = \frac{2}{9}x^{2}$

$y = \frac{2}{9}x^{2}$

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Example Question #511 : Calculus Ii

Write in Cartesian form:

x = 2t - 1

y = 4t+2

Possible Answers:

$y = \frac{1}{2}x + 2$

y = 2x - 4

y = 2x

y = 2x + 4

$y = \frac{1}{2}x - 2$

Correct answer:

y = 2x + 4

Explanation:

y = 4t+2 = 4t - 2 + 4 = 2 (2t - 1)+ 4 = 2x + 4 ,

so the Cartesian equation is

y = 2x + 4 .

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Example Question #4 : Parametric

Write in Cartesian form:

$x = \ln (t+1)$

y = t^2 + 2t + 2

$t \in (-1, \infty )$

Possible Answers:

$y = e^{2x} + 1$

$y = e^{2x} + e$

$y = e^{2x} - e$

$y = e^{2x}+ 2$

$y = e^{2x} - 1$

Correct answer:

$y = e^{2x} + 1$

Explanation:

$x = \ln (t+1)$

$e ^{x} = \left [ \ln (t+1) \right ] ^{x}$

$e ^{x} = t + 1$

$y = t^2 + 2t + 2 = t^2 + 2t + 1 + 1 = (t + 1)^{2}+ 1 = \left (e^{x} \right )^{2} + 1 = e^{2x} + 1$

Therefore the Cartesian equation is $y = e^{2x} + 1$ .

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Example Question #2 : Parametric, Polar, And Vector Functions

Rewrite as a Cartesian equation:

$x =10^ {t}, y = \sinh t, t \in (0,\infty )$

Possible Answers:

$y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

$y = \frac{ x ^{2 \log e} -1}{ x ^{2 \log e} + 1}$

$y = \frac{ x ^{2 \log e} }{2 x ^{\log e}}$

$y = \frac{ x ^{2 \log e} +1}{2 x ^{\log e}}$

$y = \frac{ 2x ^{2 \log e} -1}{2 x ^{\log e}}$

Correct answer:

$y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

Explanation:

$y = \sinh t = \frac{e ^{2t} -1}{2e ^{t}} = \frac{\left (e ^{ t} \right) ^{2} -1}{2e ^{t}}$

$e^{t} = 10 ^{\log e \cdot t} =\left ( 10 ^{ t} \right )^{\log e} = x ^{\log e}$ , so

$y = \frac{\left ( x ^{\log e} \right) ^{2} -1}{2 x ^{\log e}}= \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

This makes the Cartesian equation

$y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$ .

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Example Question #7 : Parametric

$\small x=3t+4$ and $\small y=1/t$ . What is $\small y$ in terms of $\small x$ (rectangular form)?

Possible Answers:

$\small y=(x-3)/4$

$\small y=4/(x-3)$

$\small y=(x-4)/3$

$\small y=3/(x-4)$

Correct answer:

$\small y=3/(x-4)$

Explanation:

In order to solve this, we must isolate $\small t$ in both equations.

$\small \small x=3t+4\rightarrow t=(x-4)/3$ and

$\small y=1/t\rightarrow t=1/y$ .

Now we can set the right side of those two equations equal to each other since they both equal $\small t$ .

$\small (x-4)/3=1/y$ .

By multiplying both sides by $\small 3y/(x-4)$ , we get $\small y=3/(x-4)$ , which is our equation in rectangular form.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #16 : Limits And Continuity

Example Question #511 : Limits

Example Question #511 : Calculus Ii

Example Question #1 : Parametric, Polar, And Vector Functions

Example Question #1 : Parametric

Example Question #2 : Parametric

Example Question #511 : Calculus Ii

Example Question #4 : Parametric

Example Question #2 : Parametric, Polar, And Vector Functions

Example Question #7 : Parametric

All Calculus 2 Resources

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