# Mathematics Graphic

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**1451** - Category: Graphic Mathematics

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Please Show all work. ( calculus 1 )

1) Show that the given families of curves are orthogonal trajectories of each other. Sketch both families of curves on the same axes.

x^{2} + y^{2} = r^{2}, ax+ by= 0

The two equations are orthogonal trajectories of each other (black circles for x^{2} + y^{2} = r^{2}), and the red lines are the family of ax+by=0. You can see that any pair can be rotated along the axis with no change of shape.

2) differentiate the function.

a) f(x)= x^{(1/2)} ln x

Using the derivative of products d(u*v)= udv + vd

We let u=x^{(1/2)} and v = lnx

b) y= ln(x^{4}Sin^{2}x)

let u = x^{4}Sin^{2}x so that y becomes y=ln(u) and applying the derivative of product for d(u)

3) Find y` and y“.

y= x ln x

Using the derivative of products d(u*v)= udv + vdu

We let u=x and v = lnx

Solving for the 1^{st} derivative y’

Solving for the 2^{nd} derivative y’’ from y’

4) Find an equation of the tangent line to the curve at the given point.

y= ln ln x, (e , 0)

Solving for the slope of the equation at any point m

we get the derivative using d(lnu)=(1/u)du where u=lnx

m = y’(x)

The slope of the tangent line mt is

mt = m

Then we evaluate the value of the slope at x = e

We get mt = 1/e

Using the point slope form y = m(x-x1) + y1 we get the equation of the tangent line

y = mt (x-x1) + y1 where x1=e and y1 = 0 we get the final answer

y = (1/e) (x – e)

5) Find the first and second derivatives of the function.

y= cos(q)

The 1^{st} derivative

y’=-sin(q)

The 2^{nd} derivative

y’=-cos(q)

6) Find y“`

y= (2x + 3 )^{1/2 }

Applying du^{n}=n*u^{n-1} where u=2x+3

y’=(1/2)*(2x+3)^{-1/2}(2)

y’’=(-1/2)*(2x+3)^{-3/2}(2)

y’’’=-(-3/2)*(2x+3)^{-5/2}(2)

y’’’=3(2x+3)^{-5/2}

7) If a snowball melts so that its surface area decreases at a rate of 1cm^2/min, find the rate at which the diameter decreases when the diameter is 10cm.

Since the equation of Surface Area (S) as a function of diameter (d) is

S=p*d^{2}

We get the derivative of both sides with respect to dt

Simplifying the equation by using rS for the rate of change of surface and using the given

We can solve for the rate of change of diameter (negative meaning decrease)

8) Find the critical numbers of the function.

s(t) = 3t^{4} + 4t^{3} – 6t^{2}

The critical numbers are found by getting the derivative and equating this to zero

s`(t) = 12t^{3}+12t^{2}-12t

t^{3}+t^{2}-t = 0

t(t^{2}+t-1) = 0

The critical numbers are

t0 = 0

9) Find the absolute max and absolute min values of f on the given interval.

Solution: Get the derivative, equate to zero, solve for x, then get f(x)

a) f(x) = 3x^{2} – 12x + 5, (0,3)

0 = 6x –12

x = 2

f(2) = 3*4 – 12*2 + 5 = -7

b) f(x) = 2x^{3} – 3x^{2} – 12x + 1, ( -2, 3)

0 = 6*x^{2} – 6x – 12

0 = x^{2} –x – 2

0 = (x-2)(x+1)

x1=2

x2=-1

f(x1) = 2*8-3*4-12*2+1

f(x1) = 16 –12*3 + 1

f(x1) = -19

f(x2) = 2*(-1)-3*(1)+12+1

f(x2) = -2-3+12+1 = 8

c) f(x) = ( x^{2} – 1)^{3}, (-1, 2)

0 = 3*(x^{2}-1)^{2} (2x)

0 = 6x(x^{2}-1)^{2}

x1 = 0

x2 = 1

x3 = -1

f(x1) = 1

f(x2) = 0

f(x3) = 0

d) f(x) = x/ (x^{2} + 1), ( 0, 2)

f(x) = x * (x^{2}+1)^{-1}

0 = – x (x^{2}+1)^{-2}(2x) + (x^{2}+1)^{-1}

0 = -2x^{2}(x^{2}+1)^{-2}+ (x^{2}+1)^{-1}

0 = -2x^{2}(x^{2}+1)^{-1}+ 1

0 = -2x^{2}+(x^{2}+1)

0 = -x^{2}+1

x = (-1)^{1/2 }

f(x) = imaginary

d) f(x) = ( ln x )/x, (1,3)

0 = -(lnx)x^{-2}+x^{-1}* x^{-1}

0 = 1 – ln x

x = e

f(x) = 1/e

10) Find the most general antiderivative of the function. ( check your answer by differentiation).

Solution by integration. C denotes a constant

a) f(x) = 10/x^{9}

f(x)=10 x^{-9}

F(x) = (-10/8)x^{-8 }+ C

b) f(x) = 6 (x)^{1/2} – (x)^{1/6}

^{ }F(x) = 6*(2/3)x^{3/2} – (6/7)x^{7/6} + C

11) If 1200 cm^{2} of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

Solution:

Let x be the width of the square box and y the height so the total area of open top considering 5 sides

1200 = x^{2}+4xy

y = (x^{2}-1200)/4x

y’ = -(x^{2}-1200)(4x)^{-2} + (4x)^{-1}(2x)

y’ = – (x^{2}-1200) + 8x^{2}

y’ = 7 x^{2 }+1200

0 = 7 x^{2 }+1200

x = 1200/7

x = 171.43 cm

y = 41.11 cm

largest volumen v

v = x*x*y

v = 1208150.75 cm^{3}

12) Write the composite function in the form f(g(x) ). { Identify the inner function u = g(x) and the outer function y= f(u).} Then find the derivative dy/dx.

y= (4 + 3x)^{1/2}

^{ }let u = 4 + 3x

y = u^{1/2}

dy = (1/2)*u^{-1/2}du

dy = (1/2)*(4+3x)^{ -1/2}(3dx)

dy/dx = (3/2)*(4+3x)^{ -1/2}

13) Find the derivative of the function.

a) f(t) = (1 + tan t)^{1/3}

Solution;

D_{t}f(t) = (1/3) (1 + tan t)^{-2/3}(sec^{2}t)

b) y= tan^{2}(3q)

Solution:

dy/dq=2tan(3q)(3)

dy/dq=6tan(3q)

14) Find the most general antiderivative of the function. ( check your answer by differentiation).

a) f(x) = x^{20} + 4x^{10} + 8

Solution

A_{x}f(x)=(1/21) x^{21} + (4/11)x^{11} + 8x + C

b) f(x) = 2x + 3x^{7}

Solution:

A_{x}f(x)=(2/2)x^{2}+(3/2.7)x^{2.7}+C

A_{x}f(x)= x^{2}+(3/2.7)x^{2.7}+C

c) f(x) = (x^{3})^{1/4} + (x^{4})^{1/3}

Solution:

f(x) = x^{3/4} + x^{4/3}

A_{x}f(x) = (4/7) x^{7/4} + (3/7)x^{7/3}+C

d) f(u) = u^4 + 3(u)^1/2 /u^2

15) Find f.

f“`(x) = 2 – 12x, f(0) = 9, f(2) = 15

Solution:

1^{st} Antiderivative of f’’’(x)

f’’(x)=2x – (12/2)x^{2} + C

f’’(x)=2x – x^{2 }+ C

2^{nd} Antiderivative

f’(x)= (2/2) x^{2 }– (1/3) x^{3 }+ Cx + C2

f’(x)= x^{2 }– (1/3) x^{3 }+ Cx + C2

3^{rd} Antiderivative

f(x) = (1/3)x^{3 }– (1/12) x^{4 }+ (C/2)x^{2} + C2x + C3 let (C/2)=C1

f(x) = (1/3)x^{3 }– (1/12) x^{4 }+ C1x^{2} + C2x + C3

f(0) = 9 = C3

f(2) = (1/3)2^{3 }– (1/12) 2^{4 }+ C1*2^{2} + C2*2 + 9 = 15

15 = (8/3) – (16/12) + 4*C1 + 2*C2

**No Solution: requires additional given f(x) to solve**

16) Given that the graph of f passes through the point (1,6) and that the slope of its tangent line at ( x, f(x) ) is 2x + 1, find f(2).

Solution :

The slope is the 1^{st} derivative

f’(x)=2x+1

1^{st} Antiderivative

f(x) = x^{2} + x + C

Using the intersection to solve for C

6 = f(1) = 1 + 1 + C

C = 4

We get the final equation f(x)

f(x) = x^{2} + x + 4

So that

f(2) = 4 + 2 + 4

f(2) = 10

17) Find the differential of the function.

a) y = cos(px)

dy = -sin(px) pdx

dy = -psin(px)dx

b) y = x ln x

c) y = (1+ t^{2})^{1/2}

dy = (1/2) (1+ t^{2})^{-1/2}(2tdt)

dy = t(1+ t^{2})^{-1/2 }dt

18) Use Part 2 of the Fundamental Theorem of Calculus to evaluate the integral, or explain why it does not exist.

a) The integration of 6 dx between 5 and -2

b) The integration of (1 + 3y – y^{2}) dy between 4 and 0

c) The integration of x^{4/5} dx between 1 and 0

d) The integration of (3/ t^{4} )dt between 2 and 1

e) The integration of cos(q)dq between 2p and p

19) Find a definition of `tangent` in a dictionary. Is it correct? Other comments?

**From Wordweb**

*A straight line or plane that touches a curve or curved surface at a point but does not intersect it at that point*

No this not entirely correct. It requires a mathematical description such as a line with the same slope as the curve at the point of intersection.