Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 4

Differentiation

Learning from Mistakes

There are mistakes in each of the following solutions. Identify the errors, and give the correct answer.

1. The displacement (in km) of a moving car is given by $s\left(t\right)=3t^3+4t-2$, where $t$ is measured in hours.
   1. Give the average velocity in the time period $\left[1,5\right]\text{.}$
   2. Give the two different interpretations of the value found in part (a), above.
   3. Give the average velocity in the time period $\left[1,1+h\right]$, for $h>0\text{.}$

   Erroneous Solution

   1. ${\displaystyle \frac{s\left(5\right)-s\left(1\right)}{1-5}=\frac{3\left(75\right)+4\left(5\right)-2-\left(3+4-2\right)}{1-5}=\frac{238}{-4}=-59\text{.}5\text{.}}$

   2. The velocity is decreasing at a rate of $59\text{.}5$ km/h, and the slope of the secant line passing through $\left(1\text{, }5\right)\text{ and }\left(5\text{, }245\right)\text{ is }59.5\text{.}$

   3. ${\displaystyle \frac{s\left(1\right)-s\left(1+h\right)}{h}=\frac{5-3{\left(1+h\right)}^3+4\left(1+h\right)-2}{h}=\frac{4-9h^2-5h-3h^3}{h}\text{.}}$

2. Use Definition 4.5 to find ${\displaystyle \frac{d}{dx}\frac{3}{x}\text{.}}$

   Erroneous Solution

   ${\displaystyle \underset{h\to 0}{\lim }\frac{3{\left(x+h\right)}^{-1}-3\left(x^{-1}\right)}{h}=\underset{h\to 0}{lim}\frac{3\left(x+h\right)-3x}{x\left(x+h\right)h}=\frac{3}{x\left(x+h\right)}=\frac{3}{x^2}\text{.}}$

3. Consider the piecewise function

   $f\left(s\right)=\left\{\begin{array}{cc}\left|x\right|\text{for}\hfill & x\le 4\hfill \\ {\left(x-6\right)}^2\text{for}\hfill & x>4\hfill \end{array}\right.$

   1. Sketch the graph of the function $f\text{.}$
   2. Sketch the graph of the derivative function $f^{\prime }\text{.}$

   Erroneous Solution

   1.  

      

      Figure 4.23. Erroneous solution, “Learning from Mistakes,” Problem 3(a)

   2.  

      

      Figure 4.24. Erroneous solution, “Learning from Mistakes,” Problem 3(b)

4. Find the derivatives indicated below.
   1. ${\displaystyle f\left(x\right)=\frac{3x^2+6x-3}{x-4x^3}}$ find $f^{\prime }\left(x\right)\text{.}$
   2. ${\displaystyle \frac{d}{dx}\sqrt{x^{3}sin x}\text{.}}$
   3. ${\displaystyle \frac{d}{dx} \; sec\sqrt{x+1}\text{.}}$
   4. ${\displaystyle \frac{d^2}{d{x}^2} tan\left(3x\right)\text{.}}$

   Erroneous Solution

   1. ${\displaystyle f^{\prime }\left(x\right)=\frac{\left(3x^2+6x-3\right)\left(1-12x^2\right)-\left(6x+6\right)\left(x-4x^3\right)}{{\left(x-4x^3\right)}^2}\text{.}}$
   2. ${\displaystyle \frac{d}{dx}\sqrt{x^{3}sin x}=\frac{3x^{2}cos x}{2\sqrt{x^{3}sin x}}}$.
   3. ${\displaystyle \frac{d}{dx} sec\sqrt{x+1}=sec x tan x\left(\frac{1}{2\sqrt{x+1}}\right)}$.
   4. ${\displaystyle \frac{d}{dx}tan\left(3x\right)={sec}^2\left(3x\right)}$   and   ${\displaystyle \frac{d}{dx}{sec}^2\left(3x\right)=2{ sec}^2\left(3x\right) tan\left(3x\right)}$.

5. A particle is moving along the curve $y=\sqrt{x}$. As the particle passes through the point $\left(4,2\right)$, its $x$-coordinate increases at a rate of $3$ cm/s. How fast is the distance from the particle to the origin changing at this instant?

   Erroneous Solution

   We have ${\displaystyle \frac{dx}{dt}=3}$, and we want to know ${\displaystyle \frac{dy}{dt}}$ when ${\displaystyle x=4}$, where $y$ is the distance from the origin to the point $\left(x,\sqrt{x}\right)$ on the curve.

   Hence, $y=\sqrt{x^2+x}\text{.}$

   Differentiating, we obtain

   ${\displaystyle \frac{dy}{dt}=\frac{2x+1}{2\sqrt{x^2+1}}\text{.}}$

   For $x=4\text{,}$

   ${\displaystyle \frac{dy}{dt}=\frac{9}{2\sqrt{17}}\text{.}}$

   The distance is increasing at a rate of ${\displaystyle \frac{9}{2\sqrt{17}}}$  cm/s.

6. A plane flying horizontally at an altitude of 1 mi and a speed of 500 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when the plane is 2 mi away from the station.

   Erroneous Solution

   We want ${\displaystyle \frac{dy}{dt}}$ when $y=2$ mi.

   

   Figure 4.25. Erroneous solution, “Learning from Mistakes,” Problem 6

   From the figure, $y^2=x^2+1$. Differentiating yields ${\displaystyle 2y\frac{dy}{dt}=2x\text{.}}$

   Then ${\displaystyle \frac{dy}{dt}=\frac{x}{y}\text{,}}$ when $y=2$

   ${\displaystyle \frac{dy}{dt}=\frac{500}{2}=250\text{ mi/h}\text{.}}$