Calculus

Dear Questioner its not able to solve two variables for one equation, give one more equation. Thanks

5

In derivative questions, the value of zero is often used as a reference point for evaluating the rate of change of a function at a specific point. By setting the change in the variable (Δx) to zero, we can analyze the instantaneous rate of change at that point, which is the essence of the derivative. This approach allows us to find the slope of the tangent line to the curve at a given point, providing important information about the function's behavior.

2x - 3y = 2

3x + 2y = 3

Simultaneous Equations.

Arbitrarily decide which letter to eliminate.

In this case 'y'.

To do this we bring the coefficients of 'y' to the same value. . To do this multiply the top equation by '2' and the bottom eq'n by.3.

Hence

4x - 6y = 4

9x + 6y = 9

We now add the two equations. The addition of '-6y + 6y = 0 ', So 'y' is eliminated.

Hence

13x = 13

x = 1

To find 'y' , we substitute 'x = 1' into either equation.

Hence

4(1) - 6y = 4

4 - 6y = 4

-6y = 0

y = 0

So the answer in ( x,y ) form is ( 1,0 ) .

-3 + 3y = 2y + 5

Add '3' to both sides

Hence

3y = 2y + 8

Subtract '2y' from both sides.

y = 8 The answer!!!!!

(3x + 2y)(5x - 3y)

Apply FOIL.

F(First) ; 3x X 5x = 15x^(2)

O(Outside) ; 3x X -3y = - 9xy

I(Inside) ; 2y X 5x = 10xy

L(Last) ; 2y X - 3y = - 6y^(2)

15x^(2) - 9xy + 10xy - 6y^(2)

Collect 'like' terms

15x^(2) + xy - 6y^(2)

The answer!!!!!!

f(x) = (1 - x^(2) ^(1/2)

Let y= (1 - x^(2)) ^(1/2)

Use Chain Rule

dy/dx = dy/du X du/dx

Let u = 1 - x^(2)

Hence y = u^(1/2)

dy/du = (1/2)u^(-1/2)

du/dx = -2x

Hence

dy/dx = dy/du X du/dx = (1/2)u^(-1/2) X ( -2x)

dy/dx = (1/2)u( 1 -x^(2)^(-1/2) X ( -2x)

Tidying up

dy/dx = (-2x/(2(1 - x^(2))^(1/2))

dy/dx = -x/ [(1 - x^(2)]^(1/2)

f(x) = 3x4 - 2x2 + 7

The function

f'(x) = 12x^(3) - 4x

!st Derivative

f''(x) = 36x - 4

2nd derivative. The answer!!!!!

To calculate the surface area to volume ratio, divide the surface area of an object by its volume (SA ÷ V). Since area scales with the square of length and volume with the cube, larger objects have a smaller surface area to volume ratio. For help with this topic, Calculus Tutoring at Smart Math Tutoring explains these concepts clearly.

\

The variable is a. You can solve for it, like this:

8a - 9 = 7a - 6

8a = 7a + 3

8a - 7a = 3

a = 3

By 'nesting'

Fifth root of 'x' is x^(1/5)

Then we 'nest' to the square root ( 1/2)

Hence

[x^(1/5]^(1/2) =

x^(1/10) or x^(0.1)

7x^(2)

= 7

Divide both sides by '7'

Hence

x^(2) = 1

Square root both sides

x = (+)1 & (-)1 , usually written as ' +/- 1 '.

Remember

(+) X (+)1 = 1

&

-1 X -1 = 1

9x9=81

9x^(2) - 6xy + y^(2) - 81

9x^(2) - 81 + y^(2) - 6xy

9(x^(2) - 9)+ y(y - 6x)

9(x-3)(x + 3) + y(y - 6x)

9x^(2)

-9x + 4 = -50

Subtract '4' from both sides

-9x = -54

Divide both sides by '-9'

x = (+)6

9x - 3 > 6

Add '3' to both sides

9x > 9

Divide both sides by '9'

x > 1 The answer!!!!!!

Verification.

When x = 2

9(2) = 18 > 9

(x^(3) + X^(2) + x + 1) / ( x + 1)

Factor

[x^(2))(x + 1) + ( x +1) ] / ) x + 1)

[(x^(2) + 1 )(x + 1)] / ( x + 1)

Cancel down by 'x + 1'

Hence

x^(2) + 1 The Answer!!!! (NB This term does NOT factor).

(1 - CosX)(1 + CosX)

Apply FOIL

Hence

F: 1 x 1 = 1

O ; 1 x CosX = CosX

I: -CosX X 1 = -CosX

L ; -CosX X CosX = - Cos^(2)x

Collect terms

1 + Cos X - CosX - Cos^(2)X

= 1 - Cos^(2)X

1 - Cos^(2)X = Sin^(2)

From the Trig/ Identity

Sin^(2)X + Cos^(2)X = 1

1-Cos^(2)x = Sin^(2)x

This is algebraically rearranged from the Trig. Identity.

Sin^(2)x + Cos^(2)x = 1

Which in turn is based on the Pythagorean triangle.

x^(2) + 8x + 8y +32 = 0

Then

-8y = x^(2) + 8x + 32

y = (-x^(2)/8) - x - 4

In standard parabolic form. The coefficient of x^(2) is '-1/8' .

It will be an 'umbrella' shaped parabola.

y = x^(2) + 4x + 5

Find the vertex , differentiate and equate to zero.

dy/dx = 0 = 2x + 4

2x + 4 - 0

2x = -4

x = -2

To find if the vertex is at a max/min differentiate are second time. If the answer is positive(+)/Negative(-), then it is a minimum/maximum.

Hence

dy/dx = 2x + 4

d2y/dx2 = (+)2 Positive(+) so the parabola is at a minimum. at x = -2.

Differential equations play a crucial role in biomedical engineering by modeling physiological processes and systems. They are used to describe dynamic behaviors such as blood flow, drug delivery kinetics, and the spread of diseases within biological tissues. Additionally, they help in the design of medical devices, such as prosthetics and imaging systems, by simulating how these devices interact with biological systems over time. Overall, differential equations provide a mathematical framework for understanding complex biological phenomena and improving healthcare solutions.

For the function y = x^(3) + 6x^(2) + 9x

Then

dy/dx = 3x^(2) + 12x + 9

At max/min dy/dx = 0

Hence

3x^(2) + 12x + 9 = 0

3(x^(2) + 4x + 3) = 0

Factor

(x + 1)(x + 3) = 0

Hence x = -1 & x = -3 are the turning point (max/min)

To determine if x = 0 at a max/min , the differentiate a second time

Hence

d2y/dx2 = 6x + 12 = 0 Are the max/min turning points.

Substitute , when x = -1

6(-1) + 12 = (+)6 minimum turning point .

x = -3

6(-3) + 12 = -6 maximum turning point.

When x = positive(+), then the curve is at a minimum.

When x = negative (-), then the curve is at a maximum turning point.

NB When d2y/dx2 = 0 is the 'point of inflexion' , where the curve goes from becoming steeper/shallower to shallower/steeper.

So when d2y/dx2 = 6x + 12 = 0

Then 6x = -12

x = -2 is the point of inflexion.

NNB When differentiating the differential answer gives the steeper of the gradient. So if you make the gradient zero ( dy/dx = 0) , there is no steepness, it is a flat horizontal line

Derivative classification authority refers to the ability granted to individuals to classify documents or information based on existing classified material. It allows authorized personnel to take previously classified information and apply classification markings to new documents that contain, derive from, or are based on that information. This authority is essential for maintaining the integrity and security of classified information while enabling the dissemination of necessary data within the bounds of national security protocols. Proper training and adherence to established guidelines are critical for individuals exercising this authority.