CoCalc -- CHAPTER 3 WS.sagews

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Display your code and results on this worksheet for these exercises:

Determine the first, second and third derivatives of the function $g(x)=e^x\sin^2 x$
Using the limit definition of derivative found in ws1, find the derivative of $\sin^2{(2x)}$ and evaluate this derivative at $x=\pi/2$ .
Create the implicit plot of the equation, $y^2=x^2+\sin xy$ , using -4 and 4 as your vertical and horizontal bounds. Then determine an x-y(x) expression for $\frac{dy}{dx}$ . Use it to find the slope(s) of any tangents to any point on the curve where $x=2$ . Create expressions for these tangent lines and plot curve, points and tangent lines on one plot.
Given the quintic function, $quin(x)=x^5/5-13x^4/4+43x^3/3+3x^2-70x+12$ , first plot it with xmin=-2 and xmax=8. Then find its first derivative. Then find any and all zeroes of its first derivative. These will be the $x$ -coordinates of all the points where the tangent is horizontal(why?). Create the ordered pairs that result in using these first derivative zeroes as the x-coordinates and add these zero-slope tangent points to the plot in black. Now do the same thing with the function's second derivative, plotting any locations of second derivative zeroes in red. We will be calling these inflection points later on.
The formula for the surface area of a cylinder is $SA=2\pi rh+2\pi r^2$ . Declare $r$ and $h$ to be functions of $t$ , then define $SA$ as a function of $t$ . Find the derivative of $SA$ .
Consider $f(x)=x^2-3x+7$ and $g(x)=-2x^3+x^2-5x-4$ . Find all x-values for which the two functions' derivatives are of equal value.

#1-- finished
1
g(x)=(e^x)*((sin(x))^2)
show(g(x))

first=(diff(g(x),x,1))
show('This is the first derivitive:',first)

second=(diff(g(x),x,2))
show('This is the second derivitive:',second)

third=(diff(g(x),x,3))
show('This is the third derivitive:',third)

1

\displaystyle e^{x} \sin\left(x\right)^{2}

This is the first derivitive:

\displaystyle 2 \, \cos\left(x\right) e^{x} \sin\left(x\right) + e^{x} \sin\left(x\right)^{2}

This is the second derivitive:

\displaystyle 2 \, \cos\left(x\right)^{2} e^{x} + 4 \, \cos\left(x\right) e^{x} \sin\left(x\right) - e^{x} \sin\left(x\right)^{2}

This is the third derivitive:

\displaystyle 6 \, \cos\left(x\right)^{2} e^{x} - 2 \, \cos\left(x\right) e^{x} \sin\left(x\right) - 5 \, e^{x} \sin\left(x\right)^{2}

#2--finished
2
h=var('h')
f(x)=(sin(2*x))^2
#now create the difference quotient function whose variable is h
diff_quo(h)=(f(x+h)-f(x))/h
#show(diff_quo)


#Now we get the derived function or derivative by taking the limit as h goes to 0!!!!!!!!
dy_dx(x)=(diff_quo(h)).limit(h=0)
show(dy_dx(x))

a=pi/2 #can edit
tangent_pt=[a,f(a)]
show(tangent_pt)
show(['the value at $\pi/2$ is',numerical_approx(f(a),digits=3),'and the instantaneous rate of change at that moment is',numerical_approx(dy_dx(a),digits=3)])

2

\displaystyle 4 \, \cos\left(2 \, x\right) \sin\left(2 \, x\right)

[

\displaystyle \frac{1}{2} \, \pi

\displaystyle 0

]

[the value at

\pi/2

is,

\displaystyle 0.000

, and the instantaneous rate of change at that moment is,

\displaystyle 0.000

]

#3 Finished!!!!!

y=var('y')
fn=(y^2==x^2+sin(x*y))
ddxeqn=(diff(fn,x))
show(ddxeqn)

new_eqn=(y^2-4-sin(2*y)) #when x=2
sols=(solve(new_eqn==0,y))
show(sols)

pt1=[2,-2.2288422554413483]
pt2=[2,1.8588338631928305]

dy_dx(x)=(diff(new_eqn)).limit(h=0)
show(dy_dx(x))
thang=(dy_dx(2))
  #can edit
#show(thang)


#y1_1=find_root(new_eqn, #1BOUND, #2BOUND)
p=implicit_plot(fn,(x,-4,4),(y,-4,4))
#p

m1=(-2.2288422554413483)*cos(2*-2.2288422554413483)+2*2
m2=(1.8588338631928305)*cos(2*1.8588338631928305)+2*2
tan_1=m1*(x-2)-2.2288422554413483
tan_2=m2*(x-2)+1.8588338631928305

p+=plot(tan_1,x,xmin=-4,xmax=4,ymin=-4,ymax=4,color='red')
p+=plot(tan_2,x,-4,4,color='green')
p+=point([pt1,pt2],size=25,color='black')
p

\displaystyle 0 = y \cos\left(x y\right) + 2 \, x

[

\displaystyle y = -\sqrt{\sin\left(2 \, y\right) + 4}

\displaystyle y = \sqrt{\sin\left(2 \, y\right) + 4}

]

\displaystyle 2 \, y - 2 \, \cos\left(2 \, y\right)

#4 FINISHED
4
f(x)=(x^5/5)-((13*x^4)/4)+((43*x^3)/3)+(3*x^2)-70*x+12
show(f(x))

first(x)=(diff(f(x),x,1))
show('this is the first derivative:',first(x))

sols=(solve(first(x)==0,x))
show(sols)

num1=sols[0].rhs()
num2=sols[1].rhs()
num3=sols[2].rhs()
num4=sols[3].rhs()

p=plot(f(x),xmin=-2,xmax=8)

pt1=(num1,f(num1))
pt2=(num2,f(num2))
pt3=(num3,f(num3))
pt4=(num4,f(num4))

p+=point2d(pt1,size=50,color='black')
p+=point2d(pt2,size=50,color='black')
p+=point2d(pt3,size=50,color='black')
p+=point2d(pt4,size=50,color='black')
p

#Now do the same thing with the function's second derivative, plotting any locations of second derivative zeroes in red.
       #We will be calling these inflection points later on.

second(x)=(diff(first(x),x))
show('this is the second derivative',second(x))


show('since the zeros for the second derivative are imaginary, they cannot be plotted on the second derivative graph')

4

\displaystyle \frac{1}{5} \, x^{5} - \frac{13}{4} \, x^{4} + \frac{43}{3} \, x^{3} + 3 \, x^{2} - 70 \, x + 12

this is the first derivative:

\displaystyle x^{4} - 13 \, x^{3} + 43 \, x^{2} + 6 \, x - 70

[

\displaystyle x = -\sqrt{6} + 4

\displaystyle x = \sqrt{6} + 4

\displaystyle x = -\frac{1}{2} \, \sqrt{53} + \frac{5}{2}

\displaystyle x = \frac{1}{2} \, \sqrt{53} + \frac{5}{2}

]

this is the second derivative

\displaystyle 4 \, x^{3} - 39 \, x^{2} + 86 \, x + 6

since the zeros for the second derivative are imaginary, they cannot be plotted on the second derivative graph

#5-- finished

5


#The formula for the surface area of a cylinder is SA=2\pi rh+2\pi r^2SA=2πrh+2πr
#Declare rr and hh to be functions of tt, then define SASA as a function of tt. Find the derivative of SASA.
r,h=var('r,h')
r=function('r')(t)
h=function('h')(t)
show(r)
show(h)

#r(t)=a
#h(t)=b
SA(t)=(2*pi*r*h)+(2*pi*r^2)
show(SA(t))

der=(diff(SA(t),t))

show('the derivative of SA(t) is',der)

5

\displaystyle r\left(t\right)

\displaystyle h\left(t\right)

\displaystyle 2 \, \pi h\left(t\right) r\left(t\right) + 2 \, \pi r\left(t\right)^{2}

the derivative of SA(t) is

\displaystyle 2 \, \pi r\left(t\right) \frac{\partial}{\partial t}h\left(t\right) + 2 \, \pi h\left(t\right) \frac{\partial}{\partial t}r\left(t\right) + 4 \, \pi r\left(t\right) \frac{\partial}{\partial t}r\left(t\right)

#6-- FINISHED
6
f(x)=x^2-3*x+7
g(x)=-2*x^3+x^2-5*x-4

show('f(x)=',f(x))
show('g(x)=',g(x))

fder=diff(f(x),x,1)
gder=diff(g(x),x,1)

show('f(x) derivative:',fder)
show('g(x) derivative:',gder)


sols=solve(fder==gder,x)
show(sols)

f(x)=

\displaystyle x^{2} - 3 \, x + 7

g(x)=

\displaystyle -2 \, x^{3} + x^{2} - 5 \, x - 4

f(x) derivative:

\displaystyle 2 \, x - 3

g(x) derivative:

\displaystyle -6 \, x^{2} + 2 \, x - 5

[

\displaystyle x = -\frac{1}{3} i \, \sqrt{3}

\displaystyle x = \frac{1}{3} i \, \sqrt{3}

]

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