maximum and perpendicular problem

Question

so am i finding the points or just the highest point which would be x? right. please show me how to do this. and i need to check my answer for number 7

Answer 1

6) s(t) = (-1/2)t² + 3t + 5

Apply derivative on each side with respect of t.

ds/dt = - t + 3

Maximum height (ds/dt) = 0

- t + 3 = 0

t = 3

The object reach maximum height in 3 seconds.

s = (-1/2)(3)² + 3(3) + 5

s = - 9/2 + 9 + 5

s = ( - 9 + 18 + 10)/2

s = 19/2

s = 9.5

Maximum height of the object is 9.5 feet.

Comments

Maximum height (ds/dt) = 0

 

how do you determine that the height is 0

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At maximum height, velocity = 0

The rate of change in displacement is called velocity.It is denoted by 'v'.

v = ds/dt  = 0

Answer 2

7)

Let the points are (x₁, y₁) = (9, 7) and (x₂, y₂) = (11, 4).

Slope (m) = [(y₂ - y₁)/(x₂ -x₁)]

m = [(4 - 7)/(11 - 9]

⇒ m = - (3/2)

We know that perpendicular lines slopes are negative reciprocal to each other.

Slope of perpendicular line (m) = 2/3

Slope-intercept form line equation is y = mx + b, where m is slope and b is y-intercept.

Now, the line equation is y = (2/3)x + b.

Find the y - intercept by substituting any point in the line equation say (x, y) = (4, 2).

2 = (2/3)(4) + b

b = 2 - (8/3)

b = (6 - 8)/3

⇒ b = - (2/3).

The line equation in slope - intercept - form is y = (2/3)x - (2/3).