Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #511 : Calculus Ii

Screen shot 2015 08 19 at 5.23.24 pm

Given the above graph of , over which of the following intervals is  continuous?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

1.) The point  must exist, and

2.) .

 

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

Example Question #511 : Limits

Screen shot 2015 08 19 at 5.43.00 pm

Given the above graph of , over which of the following intervals is  continuous?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

1.) The point  must exist, and

2.) .

 

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

Example Question #12 : Limits And Continuity

Screen shot 2015 08 21 at 11.18.01 am

Given the above graph of , over which of the following intervals is continuous?

Possible Answers:

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

\displaystyle (-\infty,0)

\displaystyle (0,\infty)

\displaystyle (1,\infty)

Correct answer:

\displaystyle (0,\infty)

Explanation:

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

1.) The point must exist, and

2.) .

 

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

Example Question #1 : Parametric Form

Rewrite as a Cartesian equation:

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

Possible Answers:

\displaystyle y = x+2

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

\displaystyle y = x - 2

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

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

Correct answer:

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

Explanation:

\displaystyle x = t^{2} + 2t + 1

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

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

So 

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

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

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

Also,

\displaystyle y = t^{2} - 2t + 1

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

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

So 

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

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

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

or equivalently,

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

to make \displaystyle t - 1 nonpositive.

 

Then,

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

and 

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

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

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

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

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

Example Question #1 : Parametric, Polar, And Vector

Write in Cartesian form:

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

Possible Answers:

\displaystyle x + 2y^{2} = 1

\displaystyle x - 2y = 1

\displaystyle x + 2y = 1

\displaystyle x - 2y^{2} = 1

\displaystyle x = 2y^{2}

Correct answer:

\displaystyle x + 2y^{2} = 1

Explanation:

Rewrite \displaystyle x using the double-angle formula:

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

Then 

\displaystyle x = 1 - 2y^{2}

\displaystyle x + 2y^{2} = 1

which is the correct choice.

Example Question #1 : Parametric

Write in Cartesian form:

\displaystyle x = 3t, y = 2t^{2}

Possible Answers:

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

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

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

\displaystyle x^{2}+ y^{2} = 36

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

Correct answer:

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

Explanation:

\displaystyle x = 3t, so 

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

 

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

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

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

Example Question #3 : Parametric, Polar, And Vector

Write in Cartesian form:

\displaystyle x = 2t - 1

\displaystyle y = 4t+2

Possible Answers:

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

\displaystyle y = 2x

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

\displaystyle y = 2x - 4

\displaystyle y = 2x + 4

Correct answer:

\displaystyle y = 2x + 4

Explanation:

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

so the Cartesian equation is 

\displaystyle y = 2x + 4.

Example Question #4 : Parametric, Polar, And Vector

Write in Cartesian form:

\displaystyle x = \ln (t+1)

\displaystyle y = t^2 + 2t + 2

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

Possible Answers:

\displaystyle y = e^{2x}+ 2

\displaystyle y = e^{2x} + 1

\displaystyle y = e^{2x} - e

\displaystyle y = e^{2x} - 1

\displaystyle y = e^{2x} + e

Correct answer:

\displaystyle y = e^{2x} + 1

Explanation:

\displaystyle x = \ln (t+1)

so 

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

\displaystyle e ^{x} = t + 1

 

\displaystyle 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  \displaystyle y = e^{2x} + 1.

Example Question #2 : Functions, Graphs, And Limits

Rewrite as a Cartesian equation:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

\displaystyle 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

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

Example Question #2 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

\displaystyle \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 \displaystyle \small t.

 \displaystyle \small (x-4)/3=1/y.

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

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