Scribe Post: Day 21 - The derivative function

Hey Everybody! It's Ashlynn here, the scribbler for today.


Todays class started off with these questions...

1) A ball is thrown up in the air with a velocity of 96 ft/sec from a height of 112 feet. The height of the ball for t>0 is given by
h(t) = -16t² + 96t + 112
a) Find the maximum height the ball attains
The equation is a parabola so we find the x-coordinate of the vertex using:
x = -b / 2a
= -96/ 2(-16)
x = 3
We found that, then we can find the y- coordinate by plugging 3 into the equation
h(3) = -16(3)² + 96(3) +112
= 256
therefore the maximum height is 256 ft.

b) Find the average velocity between t=0 and t=7.

Avg. velocity =(h(7) - h(0) ) / (7-0)
= 0 - 112 / 7
= -16 ft./sec

c) Find the velocity at t=3

There were a number of ways Mr. K showed us on how to answer this question.

i) h(3.001) - h(3) / (3.001 - 3)
= -.016

ii) Was to use your slope program in your calulator

iii) lim g(x+h) - g(x) / h
h->0

lim [ (-16(x+h)² + 96(x+h) + 112) - h(t) ] / h
h->0

If you expand it you find the derivative fuction that help you determine the
velocity at any point.

h ' (t) = -32t +96
h ' (3) = -32(3) +96
= 0
h ' (3)= 0 so the slope of the tangent at t=3 is 0.

iv) Find the average velocity between two estimates.

lim h(3.001) - h(3) / (3.001 - 3)
h->0 = 0.016

lim h(3) - h(2.999) / (3 - 2.999)
h->0 = 0.016

( 0.016 +0.016 ) / 2
= 0.016

v) use your calculator to draw the tangent at x=3

1) [2nd] Draw
2) 5:tangent
3) x = 3
4) [enter]

You will find that the slope of the tangent line is 0.


d) Estimate, to the nearest hundredth, the velocity of the ball at the instant it hits
the ground.

To solve this we use the Quadratic formula

t= -b +/- squareroot ( b² - 4ac) / 2a

t = -96 + squareroot ( 96² - 4(-16)(112) / 2(-16)
= 7

t = -96 - squareroot ( 96² - 4(-16)(112) / 2(-16)
= -1
* t>0 so we can discard t = -1.*


2) A function f has a derivative f '(x) = x³ - 3x +1. Use your calculator to sketch f'.
a) Where on the graph of f' are the tangent lines horizontal?

I would include a graph but, I can't put the image on here. So, you can see it in
your calculator

On the graph you would look at the roots because it is a derivative function of f.
The roots are where the slope of the tangent lines would equal zero.

The roots are x = -1.8794, 0.3473, and 1.5320.

b) Suppose f(-1)=5. Write the equation of the tangent line to f at x=-1.

f(-1)=5 , x=-1 (-1,5) <---- point

f '(-1)= (-1)³ - 3(-1) +1
f '(-1)= 3 <---- Slope

We have a point and a slope, so we use the point-slope formula and use the point
(-1,5) and slope of 3.
y - y₁ = m (x - x₁)

y - 5 = 3 (x +1)
* We can leave the equation as it is *

3) a) f(2)=3 and f '(2)=4. Write an equation for the line tangent to f at x=2.

f ' (2) =4 <--- Slope
The method is the same as question 2 b.
(2,3) and slope of for f is 4.

y - y₁ = m (x - x₁)
y - 3 = 4 (x-2)


b) f(x)=x² and f'(x)=2x. Write an equation for the tangent line to f
at x=3.

f(3) = (3)²
= 9

f ' (3) = 2(3)
= 6

y - 9 = 6(x - 3)

c) Suppose f(x)= 2e^3xFind an equation for the tangent line to f
at x=0.

f(0) = 2e³⁽⁰⁾
= 2

To find the slope we can use our calculator to find the derivative at 0:

1) hit [math]
2) 8:nDeriv(
3) (y₁, x, 0)
4) [enter]

you will find that the slope is 6.000009

we now have a slope, 6, and a point (0,2)

y - 2 = 6(x-0)---> y - 2 = 6x


Homework for tonight is section 2.4 odd questions + 10 & 14. The next scibe is going to be done by... Crystal.