Limits - calculus

Question

An object is thrown straight up. Its height is given by

y(t) = 100t − 16t2

with y in feet and t in seconds. Write a formula for the average velocity on the interval

[3.5, x].

compute the limit of your formula as x approaches 3.5. You answer must be accurate to one decimal place with correct units.

lim

x→3.5 Δy/Δt = ____________

Answer 1

Given:

y(t) = 100t - 16t²

Find:

a) Average velocity formula on the closed interval [3.5, x]

b) Evaluate the limit as x approaches 3.5

Solution:

y(x) = 100x - 16x²

y(3.5) = 154

Δy/Δt = [ ( 100x - 16x² ) - 154 ] / ( x - 3.5 )

Δy/Δt = ( -16x² + 100x - 154 ) / ( x - 3.5 )

To find out if the numerator factors, use the quadratic formula on with:

a = -16, b = +100, & c = -154

[ -(100) ± √[ (100)² - 4( -16 )( -154 ) ] ] / ( 2 * -16 )

After some time and reducing to simpler numbers, you'll arrive at:

( 25 ± 3 ) / 8 = 11/8, 7/2 = 2.75, 3.5

This means, the numerator factors to:

-2 (2x - 7) (4x - 11) (if you FOIL the entire equation to the left there, you'll arrive at the equation in the numerator)

Now, I'm going to factor off another 2 from (2x - 7) [note: most limit problems, this won't typically happen] to make 2 * ( x - 3.5 )

Rewriting the average speed equation over the interval:

Δy/Δt = 4 ( x - 3.5 )( 11 - 4x ) / ( x - 3.5 )

x - 3.5 appears in numerator and denominator, simplifying the equation to:

Δy/Δt = 4 ( 11 - 4x )

a) Δy/Δt = 44 - 16x

∴ lim_x→3.5 ( Δy/Δt ) = lim_x→3.5 ( 44 - 16x ) = 44 - 16 (3.5) = 44 - 56

b) ∴ lim_x→3.5 ( Δy/Δt ) = -12

I hope this helps! Message me in the comments with any questions, comments, or concerns on what I did above! I cannot stress enough though that this was a LUCKY problem where the numerator factored so nicely like that (you won't find that for every problem, but sometimes you'll find a diamond in the ruff like this one).