1. (1 pt each) the equation of motion of a particle is s = t 3 − 12t 2 + 36t, t ≥ 0, where s is measured in meters and t is in seconds. (a) find the velocity at time t. (b) what is the velocity after 3 s? (c) when is the particle at rest? (d) when is the particle moving in the positive direction? (e) find the total distance traveled during the first 8 s. (f) find the acceleration at time t and after 3 s. 2. (3 pt each) given p(x) = x n − x, find the intervals over which p(x) is a decreasing function when: (a) n = 2 (b) n = 1 2 (c) n = −1 3. (a) (3 pts) find equations of both lines through the point (2, −3) that are tangent to the parabola y = x 2 + x. (b) (1 pt) show that there is no line through the point (2, 7) that is tangent to the parabola. then sketch a graph to see why. 4. (3 pts) find a cubic function y = ax3+bx2+cx+d whose graph has horizontal tangents at the points (−2, 6) and (2, 0). 5. (a) (3 pts) use the product rule twice to prove that if f, g, and h are differentiable, then (fgh) 0 = f 0 gh + fg0h + fgh0 . (b) (3 pts) use part (a) to differentiate y = x sin x cos x 6. (3 pts)

QUESTION 1)  

The particle's equation of motion is  

where s is measured in meters and t is in seconds.  

The velocity at time, is given by the first derivative of the equation of motion of the particle.  

Question 1b).  

To find the velocity after 3 seconds, we put into the velocity function.  

QUESTION 1C  

To find the time that the particle is at rest, we equate the velocity (the first derivative formula) to zero and solve for t.  

Divide through by 3 to get;  

Factor:  

The particle is at rest when and .  

QUESTION 1d.  

The particle is moving in a positive direction when the velocity is greater than zero.  

But .  

This implies that, the particle is moving in a positive direction on the interval,  

QUESTION 1e.  

To find the total distance traveled after 8s, we substitute into the equation of motion of the particle.  

The particle covered 32m in the first 8 seconds.  

QUESTION 1f  

i) The acceleration at time can be obtained by differentiating the velocity equation.  

ii) To find the acceleration after 3 seconds, we substitute into the equation of acceleration.  

After 3 seconds, the particle is decelerating at 6 meters per seconds square.  

QUESTION 2a  

Given:  

When , then  

This implies that, p(x) is decreasing when  

Therefore the function is decreasing on;  

QUESTION 2b  

When  

To find the interval over which the function is decreasing, we solve the inequality;  

Therefore the function is decreasing on the interval;  

QUESTION 3a

The given parabola has equation  

Let the two tangents from the external point; have equation;

Put equation (1) into equation (2)

This implies that;

Rewrite to obtain a quadratic equation in .

Since this is the point of intersection of a tangent and a parabola, the discriminant of this quadratic equation must be zero.

We substitute the values of m into equation (2) to obtain the equations of the two tangents to be;

and

QUESTION 3b

Let the equation of the tangents from the external point (2,7) be  

The given parabola has equation

The discriminant of the intersection of these two equations yields;

This quadratic equation has no real roots.

Hence there are no lines through the point (2,7) that are tangents to parabola.

We can see from the graph that this point is lying inside the parabola.

QUESTION 4

The given cubic function is  

The horizontal tangents occurs when .

This occurs at (2,0).

and (-2,6) .

These points also lie on the curve so they must satisfy the equation of the curve;

Substituting (2,0) into the original equation gives;

Substituting (-2,6) into the original equation gives;

Solving the four equations simultaneously gives;

Hence the required cubic function;

QUESTION 5a

Let  

where and are differentiable.

Using the product rule;

Use the product rule again;

as required.

answer: e 9, 5.75

step-by-step explanation:

step 1

find the variation ab in the x coordinate

delta x=(x2-x1)=10-> 8 units

find the variation ab  in the y coordinate 

delta y=(y2-y1)=6-> 2 units

step 2

find the coordinates point c

  x coordinate 

cx=ax+delta x*2/> cx=2+8*(2/> cx=3

  y coordinate 

cy=ay+delta y*2/> cy=4+2*(2/> cy=4.25

the coordinates of point c are (3,4.25)

step 3

find the coordinates point d

  x  coordinate 

dx=ax+delta x*4/> dx=2+8*(4/> dx=4

  y  coordinate 

dy=ay+delta y*4/> dy=4+2*(4/> dy=4.5

the coordinates of point d are (4, 4.5)

step 4

find the coordinates point e

  x  coordinate 

ex=ax+delta x*4/> ex=2+8*(6/> ex=5

  y  coordinate 

ey=ay+delta y*4/> ey=4+2*(6/> ey=4.75

the coordinates of point e are (5, 4.75)

step 5

find the coordinates point f

  x  coordinate 

fx=ax+delta x*4/> fx=2+8*(8/> fx=6

  y  coordinate 

fy=ay+delta y*4/> fy=4+2*(8/> fy=5

the coordinates of point f are (6,5)

step 6

find the coordinates point g

  x  coordinate 

gx=ax+delta x*4/> gx=2+8*(10/> gx=7

  y  coordinate 

gy=ay+delta y*4/> gy=4+2*(10/> gy=5.25

the coordinates of point g are (7,5.25)

step 7

find the coordinates point h

  x  coordinate 

hx=ax+delta x*4/> hx=2+8*(12/> hx=8

  y  coordinate 

hy=ay+delta y*4/> hy=4+2*(12/> hy=5.5

the coordinates of point h are (8,5.5)

step 8

find the coordinates point i

  x  coordinate 

ix=ax+delta x*4/> ix=2+8*(14/> ix=9

  y  coordinate 

iy=ay+delta y*4/> iy=4+2*(14/> iy=5.75

the coordinates of point i are (9,5.75)

- r3ktforgood ☕