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OlgaM077 [116]
2 years ago
13

An iron ball is bobbing up and down on the end of a spring. The maximum height of the ball is 46 inches and its minimum height i

s 18 inches. It takes the ball 2 seconds to go from its maximum height to its minimum height. Which model best represents the height, h, of the ball after t seconds?
Mathematics
2 answers:
LenKa [72]2 years ago
8 0

Answer:

The equation for height is

h=32sin(\frac{\pi}{2}t)+14

Step-by-step explanation:

we are given

An iron ball is bobbing up and down on the end of a spring

So, height function must be trigonometric in nature

So, we can use formula

h=Asin(Bt)+D

now, we can find A , B and D

Calculation of A:

maximum height 46 inch

minimum height =18 inch

so,

A=\frac{46+18}{2}=32

Calculation of B:

It takes the ball 2 seconds to go from its maximum height to its minimum height

So, half of time period is 2 sec

\frac{T}{2}=2

T=4

now, we can use period formula

T=\frac{2\pi}{B}

we can find B

4=\frac{2\pi}{B}

B=\frac{2\pi}{4}

B=\frac{\pi}{2}

Calculation of D:

Max=46

min=18

D=\frac{46-18}{2}

D=14

now, we can plug these values into formula

and we get

h=32sin(\frac{\pi}{2}t)+14


Nadya [2.5K]2 years ago
6 0

Answer:

Required model is  h=32\sin(\frac{\pi}{2}t)+14

Step-by-step explanation:

Given : An iron ball is bobbing up and down on the end of a spring. The maximum height of the ball is 46 inches and its minimum height is 18 inches. It takes the ball 2 seconds to go from its maximum height to its minimum height.

To find : Which model best represents the height, h, of the ball after t seconds?

Solution :  

According to question,

The height function must be trigonometric in nature.

So, we can use formula,

h=A\sin(Bt)+D

Now, We calculate A,B and D

1) Maximum height 46 inch

Minimum height =18 inch

Average height is A

A=\frac{46+18}{2}=32

2) It takes the ball 2 seconds to go from its maximum height to its minimum height

So, half of time period is 2 sec

i.e, \frac{T}{2}=2

T=4

Period is B

T=\frac{2\pi}{B}

4=\frac{2\pi}{B}

B=\frac{2\pi}{4}

B=\frac{\pi}{2}

3) D is the midline

So, Max=46  and min=18

D=\frac{46-18}{2}

D=14

Substituting all the values,

A=32 , D=14 , B=\frac{\pi}{2}

h=A\sin(Bt)+D

h=32\sin(\frac{\pi}{2}t)+14

Therefore, Required model is  h=32\sin(\frac{\pi}{2}t)+14

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Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

                                g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

                                m = f'(x) = g'(x_o)

- We will develop the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

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- For x_o = 10x + 2  ,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

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                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Solve the quadratic equation above:

                                 x = -0.0574, -0.387      

- Largest slope is at x = -0.387 where equation of line is:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- Other tangent line:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

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