Wize University Calculus 1 Textbook > Applications of Differentiation

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Position, Velocity, and AccelerationNext Section

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An object travels on a path that draws out a square which is 20 meters on a side. The object starts at corner A, and moves towards corner B. When it is halfway to corner B with a speed of 10 meters per second, at what rate is the object’s distance from corner C (the corner directly opposite corner A) changing? Do not simplify your answer.

−100500\frac{-100}{\sqrt{500}}500​−100​

100500\frac{100}{\sqrt{500}}500​100​

−10500\frac{-10}{\sqrt{500}}500​−10​

10500\frac{10}{\sqrt{500}}500​10​

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 If a tree trunk adds 1/41/41/4 of an feet to its diameter and 111 foot to its height each year, how rapidly is its volume changing when its diameter is 333 feet and its height is 505050 feet (assume that the tree trunk is a circular cylinder).

Answer

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Practice Question: Related Rates
A plane flies 3 km over a student, moving parallel to the ground at 2km/s. How fast is its distance from the student changing when it has travelled 4 km?

Answer

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Consider the function 
f(x)=sin⁡(x)f(x) = \sin(x)f(x)=sin(x)
a) Find the linear approximation of the function f(x)f(x)f(x) near x=0x = 0x=0 .

b) Use the linear approximation from part a) to approximate the value of sin⁡(0.1)\sin(0.1)sin(0.1) .

c) The approximation sin⁡(x)≈x\sin(x) \approx xsin(x)≈x is called the small angle approximation in physics. Is this name appropriate? Explain why in a sentence or two.

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The linear approximation L(x)L(x)L(x) of the function f(x)=cos⁡xf(x)=\cos{x}f(x)=cosx at x=π3x=\frac{\pi}{3}x=3π​ is 

32−12(x−π3)\frac{\sqrt{3}}{2}-\frac{1}{2}(x-\frac{\pi}{3})23​​−21​(x−3π​)

12−32(x−π3)\frac{1}{2}-\frac{\sqrt{3}}{2}(x-\frac{\pi}{3})21​−23​​(x−3π​)

32+12(x−π3)\frac{\sqrt{3}}{2}+\frac{1}{2}(x-\frac{\pi}{3})23​​+21​(x−3π​)

12+32(x−π3)\frac{1}{2}+\frac{\sqrt{3}}{2}(x-\frac{\pi}{3})21​+23​​(x−3π​)

cos⁡x−(sin⁡x)(x−π/3)\cos x-(\sin x)(x-\pi/3)cosx−(sinx)(x−π/3)

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Practice Question
Find the Taylor series representation for f(x)=11+2xf(x)=\frac{1}{1+2x}f(x)=1+2x1​ about c=−2c=-2c=−2. Then, state the radius of convergence.
  

f(x)=−∑n=0∞2n3n+1(x+2)n\displaystyle f(x)=-{\displaystyle \sum_{n=0}^\infty}
\frac{2^n}{3^{n+1}}(x+2)^nf(x)=−n=0∑∞​3n+12n​(x+2)n
Radius of convergence is R=12R=\frac{1}{2}R=21​

f(x)=−∑n=0∞2n3n+1(x+2)n\displaystyle f(x)=-{\displaystyle \sum_{n=0}^\infty}
\frac{2^n}{3^{n+1}}(x+2)^nf(x)=−n=0∑∞​3n+12n​(x+2)n
Radius of convergence is R=32R=\frac{3}{2}R=23​

f(x)=∑n=0∞2n3n(x+2)n\displaystyle f(x)={\displaystyle \sum_{n=0}^\infty}
\frac{2^n}{3^{n}}(x+2)^nf(x)=n=0∑∞​3n2n​(x+2)n
Radius of convergence is R=12R=\frac{1}{2}R=21​

f(x)=∑n=0∞2n3n(x+2)n\displaystyle f(x)={\displaystyle \sum_{n=0}^\infty}
\frac{2^n}{3^{n}}(x+2)^nf(x)=n=0∑∞​3n2n​(x+2)n
Radius of convergence is R=32R=\frac{3}{2}R=23​

f(x)=−∑n=0∞2n+13n+1(x+2)n\displaystyle f(x)=-{\displaystyle \sum_{n=0}^\infty}
\frac{2^n+1}{3^{n+1}}(x+2)^nf(x)=−n=0∑∞​3n+12n+1​(x+2)n
Radius of convergence is R=12R=\frac{1}{2}R=21​

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Find the 2nd degree Taylor polynomial of f(x)=e2xf\left(x\right)=e^{2x}f(x)=e2xcentered at a=ln⁡3a=\ln\sqrt{3}a=ln3​.

(i.e. find the first 3 terms (i.e.n=0,1,2n=0,1,2n=0,1,2) of the Taylor polynomial of f(x)f\left(x\right)f(x) with a=ln⁡3a=\ln\sqrt{3}a=ln3​)

3+(x−ln⁡3)+(x−ln⁡3)22!3+\left(x-\ln\sqrt{3}\right)+\frac{\left(x-\ln\sqrt{3}\right)^2}{2!}3+(x−ln3​)+2!(x−ln3​)2​

3+6(x−ln⁡3)+6(x−ln⁡3)22!3+6\left(x-\ln\sqrt{3}\right)+\frac{6\left(x-\ln\sqrt{3}\right)^2}{2!}3+6(x−ln3​)+2!6(x−ln3​)2​

3+6(x−ln⁡3)+6(x−ln⁡3)23+6\left(x-\ln\sqrt{3}\right)+6\left(x-\ln\sqrt{3}\right)^23+6(x−ln3​)+6(x−ln3​)2

(x−ln⁡3)+(x−ln⁡3)22!+(x−ln⁡3)33!\left(x-\ln\sqrt{3}\right)+\frac{\left(x-\ln\sqrt{3}\right)^2}{2!}+\frac{\left(x-\ln\sqrt{3}\right)^3}{3!}(x−ln3​)+2!(x−ln3​)2​+3!(x−ln3​)3​

6(x−ln⁡3)+6(x−ln⁡3)22!+24(x−ln⁡3)33!6\left(x-\ln\sqrt{3}\right)+\frac{6\left(x-\ln\sqrt{3}\right)^2}{2!}+\frac{24\left(x-\ln\sqrt{3}\right)^3}{3!}6(x−ln3​)+2!6(x−ln3​)2​+3!24(x−ln3​)3​

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Consider f(x)=1+xf(x) = \sqrt{1+x}f(x)=1+x​.
a) Find the Maclaurin polynomial of degree 3 of f(x)=1+xf(x) = \sqrt{1+x}f(x)=1+x​.
b) Provide an upper bound on the error of using T3(x)T_3(x)T3​(x) to approximate f(x)f(x)f(x) on [0,0.1][0,0.1][0,0.1].

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Which of the following points on the following function should you NOT use in order to use Newton's Method?
f(x)=x3−2x+2f(x) = x^3 - 2x + 2f(x)=x3−2x+2
Choose all that apply.

a)  23\sqrt{ \frac{2}{3} }32​​  

b)  −2-2−2 

c)  −23- \sqrt{ \frac{2}{3} }−32​​

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Use 1 iteration of Newton's method to approximate 93\sqrt[3]{9}39​ where the initial position is x=2

🌶️ TOUGH!  Evaluate the limit L=lim⁡x→∞ln⁡(e2x+ x)xL=\lim\limits_{x\rightarrow \infin}\frac{\ln(e^{2x}+\ x)}{x}L=x→∞lim​xln(e2x+ x)​.

 Evaluate the following limit  lim⁡x→∞x2+xe2x+1\displaystyle \lim_{x\rightarrow\infty}\frac{x^2+x}{e^{2x}+1}x→∞lim​e2x+1x2+x​.

Answer

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🦊 TRICKY!  Find the value of the limit lim⁡x→03x2ex\lim\limits_{x\rightarrow0}\frac{3x}{2e^x}x→0lim​2ex3x​.

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 Evaluate the limit lim⁡x→0+x3/2ln⁡x\displaystyle \lim_{x\rightarrow0^+}x^{3/2}\ln{x}x→0+lim​x3/2lnx.

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 Evaluate the following limit lim⁡x→∞xsin⁡(1x2)\displaystyle \lim_{x\rightarrow\infty}x\sin{\left(\frac{1}{x^2}\right)}x→∞lim​xsin(x21​).

Answer

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 Evaluate  
lim⁡x→3x2−9x−3\displaystyle\lim_{x\rightarrow 3} \dfrac{x^2-9}{\sqrt{x}-\sqrt{3}}x→3lim​x​−3​x2−9​

Evaluate 
lim⁡x→∞x2−10x−x2\displaystyle\lim_{x\rightarrow \infty} \sqrt{x^2-10x}-x^2x→∞lim​x2−10x​−x2

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True or False? The function f(x)=ecos⁡x+2xxf\left(x\right)=e^{\cos x}+2x^xf(x)=ecosx+2xx  has an absolute maximum and an absolute minimum in the interval [0.1]\left[0.1\right][0.1] 

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The function f(x)=1x2f(x) = \frac{1}{x^2}f(x)=x21​ has no horizontal tangent lines on the interval [−1,1][-1,1][−1,1] , even though f(−1)=f(1)f(-1)=f(1)f(−1)=f(1). Why doesn't this contradict Rolle's Theorem?

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Show that the equation cos⁡2(x)=2x\cos^2(x)=2xcos2(x)=2x has exactly one solution on R\mathbb{R}R.

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Why doesn't the function f(x)=∣x∣f(x) = \left|x\right|f(x)=∣x∣on the interval [−1,1][-1,1][−1,1]violate the Mean Value Theorem?
In other words, how is it that f(1)−f(−1)1−(−1)=0\dfrac{f(1)-f(-1)}{1-(-1)}=01−(−1)f(1)−f(−1)​=0, but there does not exist a value of cccsuch that f′(c)=0f'(c) = 0f′(c)=0?

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 For the function g(x)=(x+1)3g(x)=(x+1)^{3}g(x)=(x+1)3 on [−1,1][-1,1][−1,1], find the number ccc guaranteed by the Mean Value Theorem.

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Q.\textbf{Q.}Q. An object is moving. Its position is given by x(t)=2t3−tx(t)=2t^3-tx(t)=2t3−t. Find time ttt at which its speed is equal to the average speed over the interval [0,1][0,1][0,1].

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Suppose that fff is a differentiable function such that f′(x)≥3f'(x)\ge3f′(x)≥3  for all xxx. If f(2)=3,f(2)=3,f(2)=3, what is the minimum possible value of f(4)f(4)f(4)?

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Find the intervals on which the function f(x)=−1x+1\displaystyle f(x)=-\frac{1}{x+1}f(x)=−x+11​ is increasing or decreasing.

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Find the intervals on which the function f(x)=x3+3x2−9x+1f(x)=x^{3}+3x^{2}-9x+1f(x)=x3+3x2−9x+1 is increasing or decreasing.

Increasing: (−∞, −3)∪(1, ∞)\left(-\infty,\ -3\right)\cup\left(1,\ \infty\right)(−∞, −3)∪(1, ∞); Decreasing: (−3, 1)\left(-3,\ 1\right)(−3, 1)

Increasing: (−3,1)\left(-3,1\right)(−3,1); Decreasing: (−∞, −3)∪(1, ∞)\left(-\infty,\ -3\right)\cup\left(1,\ \infty\right)(−∞, −3)∪(1, ∞)

Increasing: (−3,1)\left(-3,1\right)(−3,1); Decreasing: (−∞,−3)∪(−3,1)∪(1,∞)\left(-\infty,-3\right)\cup\left(-3,1\right)\cup\left(1,\infty\right)(−∞,−3)∪(−3,1)∪(1,∞)

Increasing: (−∞,−3)∪(−3,1)∪(1,∞)\left(-\infty,-3\right)\cup\left(-3,1\right)\cup\left(1,\infty\right)(−∞,−3)∪(−3,1)∪(1,∞); Decreasing: (−3, 1)\left(-3,\ 1\right)(−3, 1)

Increasing: (−∞,−3)\left(-\infty,-3\right)(−∞,−3); Decreasing: (1,∞)\left(1,\infty\right)(1,∞)

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Find the value(s) of ccc such that f(x)=cx3−3(c+1)xf\left(x\right)=cx^3-3\left(c+1\right)xf(x)=cx3−3(c+1)x has

Find all the critical points of the function 

f(x) = ln⁡(x2+4x+14)f\left(x\right)\ =\ \ln\left(x^2+4x+14\right)f(x) = ln(x2+4x+14)

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Find the absolute extrema of f(x)=2xx2+1\displaystyle f(x)=\frac{2x}{x^2+1}f(x)=x2+12x​ on the interval [0,2][0,2][0,2].

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Find and classify all local and absolute extrema of f(x)=sin⁡xcos⁡xf\left(x\right)=\sin x\cos xf(x)=sinxcosx on the interval [0,  2π]\left[0,\ \ 2\pi\right][0,  2π].

x=π4 and x=5π4x=\frac{\pi}{4}\ \text{and}\ x=\frac{5\pi}{4}x=4π​ and x=45π​are local and absolute maximums;
x=3π4 and x=7π4x=\frac{3\pi}{4}\ \text{and}\ x=\frac{7\pi}{4}x=43π​ and x=47π​ are local and absolute minimums

x=π4x=\frac{\pi}{4}x=4π​is the only absolute maximum;
x=3π4x=\frac{3\pi}{4}x=43π​ is a local maximum;
x=3π4x=\frac{3\pi}{4}x=43π​ is the only absolute minimum;
x=7π4x=\frac{7\pi}{4}x=47π​ is a local minimum

x=3π4 and x=7π4x=\frac{3\pi}{4}\ \text{and}\ x=\frac{7\pi}{4}x=43π​ and x=47π​ are local and absolute maximums;
x=π4 and x=5π4x=\frac{\pi}{4}\ \text{and}\ x=\frac{5\pi}{4}x=4π​ and x=45π​are local and absolute minimums

x=π4x=\frac{\pi}{4}x=4π​is the only absolute minimum;
x=3π4x=\frac{3\pi}{4}x=43π​ is a local minimum;
x=3π4x=\frac{3\pi}{4}x=43π​ is the only absolute maximum;
x=7π4x=\frac{7\pi}{4}x=47π​ is a local maximum

x=π4x=\frac{\pi}{4}x=4π​ is the only absolute and local maximum;
x=7π4x=\frac{7\pi}{4}x=47π​ is the only absolute and local minimum

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Given the function f(x)=1x2+1f\left(x\right)=\frac{1}{x^2+1}f(x)=x2+11​, answer the following questions.

The graph of this function has one critical point, find the coordinates of this critical point.

(0, 1)\left(0,\ 1\right)(0, 1)

(1, 12)\left(1,\ \frac{1}{2}\right)(1, 21​)

(−1, 12)\left(-1,\ \frac{1}{2}\right)(−1, 21​)

(2, 15)\left(2,\ \frac{1}{5}\right)(2, 51​)

(−2. 15)\left(-2.\ \frac{1}{5}\right)(−2. 51​)

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 Graph y=xe−3xy=xe^{-3x}y=xe−3x for the domain [0,1][0,1][0,1]. Label any critical point 
 or inflection point.

12e−1\frac{1}{2}e^{-1}21​e−1, 23e−2\dfrac{2}{3}e^{-2}32​e−2

13e\frac{1}{3}e31​e, 23e2\dfrac{2}{3}e^{2}32​e2

13e−1\frac{1}{3}e^{-1}31​e−1, 23e−2\dfrac{2}{3}e^{-2}32​e−2

12e\frac{1}{2}e21​e, 23e2\dfrac{2}{3}e^{2}32​e2

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Q.\textbf{Q.}Q. Sketch of the graph of f(x)=x2−9x2+1\displaystyle f(x)=\frac{x^2-9}{x^2+1}f(x)=x2+1x2−9​. The first and the second derivatives are
f′(x)=20x(x2+1)2 and f′′(x)=20[−3x2+1(x2+1)2]\boxed{f^{\prime}(x)=\frac{20x}{(x^2+1)^2}\quad\text{ and }\quad f^{\prime\prime}(x)=20\bigg[\frac{-3x^2+1}{(x^2+1)^2}\bigg]}f′(x)=(x2+1)220x​ and f′′(x)=20[(x2+1)2−3x2+1​]​

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Sketch the graph of f(x)=xe−x2/2.f\left(x\right)=xe^{-x^2/2}.f(x)=xe−x2/2.

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Find the area of the largest rectangle that has two vertices on the  xxx-axis and two vertices lying on the graph of y=8−x2y=8-x^2y=8−x2 with 
−8≤x≤8-\sqrt{8}\leq x\leq \sqrt{8}−8​≤x≤8​.

1683+2(83)216\sqrt{\dfrac{8}{3}}+2\left(\sqrt{\dfrac{8}{3}}\right)^21638​​+2(38​​)2

1683−2(83)216\sqrt{\dfrac{8}{3}}-2\left(\sqrt{\dfrac{8}{3}}\right)^21638​​−2(38​​)2

1683+2(83)316\sqrt{\dfrac{8}{3}}+2\left(\sqrt{\dfrac{8}{3}}\right)^31638​​+2(38​​)3

1683−2(83)316\sqrt{\dfrac{8}{3}}-2\left(\sqrt{\dfrac{8}{3}}\right)^31638​​−2(38​​)3

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A cylindrical cone without a top is made to contain  81π81\pi81π cm3cm^3cm3 of liquid. Determine the dimensions of the can that minimize the area 
of metal used.

r=3;h=3r=3;h=3r=3;h=3

r=9;h=9r=9;h=9r=9;h=9

r=813;h=813r=\sqrt[3]{81}; h=\sqrt[3]{81}r=381​;h=381​

r=815;h=815r=\sqrt[5]{81};h=\sqrt[5]{81}r=581​;h=581​

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Suppose you have a rectangle of perimeter 20. Find the dimensions that will maximize the area.