Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #103 : Vector Calculations

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

Calculate \(\displaystyle \vec{v} \cdot \vec{u}\)

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Calculate the dot product of the 2 vectors.

In general,

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

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

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

Solution:

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

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

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

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

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

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

Example Question #104 : Vector Calculations

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

Calculate \(\displaystyle \vec{v} + \vec{u}\)

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Calculate the sum of vectors.

In general,

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

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

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

Solution:

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

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

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

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

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

Example Question #105 : Vector Calculations

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

Calculate \(\displaystyle \vec{v} \cdot \vec{u}\)

Possible Answers:

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

\(\displaystyle \vec{v}\cdot\vec{u}=x^2+2xe^x+4x+32\)

\(\displaystyle \vec{v}\cdot\vec{u}=2x^2+e^x+5x+32\)

\(\displaystyle \vec{v}\cdot\vec{u}=2x^2+7x+32\)

Correct answer:

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

Explanation:

Calculate the dot product of the 2 vectors.

In general,

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

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

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

Solution:

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

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

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

 

Example Question #265 : Vector

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

Calculate \(\displaystyle \vec{v} \cdot \vec{u}\)

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Calculate the dot product of the 2 vectors.

In general,

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

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

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

Solution:

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

 

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

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

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

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

Example Question #611 : Parametric, Polar, And Vector

Find the sum of the vectors \(\displaystyle u\) and \(\displaystyle v\) if

\(\displaystyle u=\left< 3, 7\right>\)

\(\displaystyle v=\left< 7, 10\right>\).

Possible Answers:

\(\displaystyle u+v=\left< 10,17\right>\)

\(\displaystyle u+v=\left< 21,70\right>\)

\(\displaystyle u+v=\left< 4,3\right>\)

\(\displaystyle u+v=\left< 17,38\right>\)

Correct answer:

\(\displaystyle u+v=\left< 10,17\right>\)

Explanation:

For the two vectors 

\(\displaystyle a=\left< a_1,a_2\right>\) and \(\displaystyle b=\left< b_1,b_2\right>\)

The sum of the two vectors a+b is 

\(\displaystyle \left< a_1, a_2\right> + \left< b_1, b_2\right>= \left< a_1+b_1, a_2+b_2\right>\)

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

\(\displaystyle \left< 3, 7\right> + \left< 7, 10\right>\)

\(\displaystyle = \left< 3+7, 7+10\right>\)

\(\displaystyle =\left < 10, 17\right>\)

Example Question #1 : Derivative Review

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

\(\displaystyle \lim_{x \rightarrow -\frac{\pi}{2}} \frac{\sin^2x -1}{x+\frac{\pi}{2}}\)

Possible Answers:

Does not exist

\(\displaystyle 0\)

\(\displaystyle \frac{1}{2}\)

\(\displaystyle 1\)

\(\displaystyle \pi\)

Correct answer:

\(\displaystyle 0\)

Explanation:

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

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

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

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

Example Question #2 : Derivative Review

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

\(\displaystyle \lim_{x \rightarrow 1} \frac{x^{49}-1}{x-1}\)

Possible Answers:

\(\displaystyle 48\)

Does not exist

\(\displaystyle 49\)

\(\displaystyle 0\)

\(\displaystyle 1\)

Correct answer:

\(\displaystyle 49\)

Explanation:

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

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

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

 

 

Example Question #1 : Derivatives

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

Possible Answers:

None of the other answers

\(\displaystyle 2\)

\(\displaystyle -8\)

\(\displaystyle 24\)

\(\displaystyle 0\)

Correct answer:

None of the other answers

Explanation:

The correct answer is 11.

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

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

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

Substituting \(\displaystyle x=1\) into both sides of the derivative we get

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

Example Question #4 : Definition Of Derivative

Evaluate the limit

\(\displaystyle \small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}\)

without using L'Hopital's rule.

Possible Answers:

\(\displaystyle \small -\frac{\sqrt{2}}{2}\)

\(\displaystyle \small 2\)

\(\displaystyle \small 1\)

\(\displaystyle \small \frac{\sqrt{2}}{2}\)

Correct answer:

\(\displaystyle \small 1\)

Explanation:

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

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

If we compare this definition to the limit 

\(\displaystyle \small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}\)

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

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

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

Example Question #1 : Derivatives

Approximate the derivative if \(\displaystyle y=\sqrt{x}\) where \(\displaystyle x=2\).

Possible Answers:

\(\displaystyle 0.35346\)

\(\displaystyle 0.35353\)

\(\displaystyle 0.35355\)

\(\displaystyle 0.35351\)

\(\displaystyle 0.35359\)

Correct answer:

\(\displaystyle 0.35355\)

Explanation:

Write the definition of the limit.

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

Substitute \(\displaystyle x=2\).

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

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

\(\displaystyle \frac{\sqrt{2+0.1}-\sqrt2}{0.1}=0.349241\)

\(\displaystyle \frac{\sqrt{2+0.0001}-\sqrt2}{0.0001}=0.353549\)

\(\displaystyle \frac{\sqrt{2+0.000001}-\sqrt2}{0.000001}=0.353553\)

The best answer is:  \(\displaystyle 0.35355\)

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