Smoke Jumpers parachute into locations to suppress forest fires

When they exit the airplane, they are in free fall until their parachutes open

https://www.youtube.com/watch?v=sovMpY2QH-I

5.5 Solving Problems Using Quadratic Relations

If the jumper exits the airplane at a height of 554m, how long will the jumper be in free fall before the parachute opens at 300m?1. Determine the quadratic relation to model

the height, H, of the smoke jumper at time t2. Determine the length of time that the

jumper is in free fall.Notice: the problem contains information about the vertex: he jumps at a height of

554m at time t = 0.Note Also: a = -0.5g = -9.8m/s2 (acceleration

due to gravity)

Smoke Jumpers – Quadratic Relations – Example #1

H = a(t – h)2 + k H = a(t – 0)2 + 554 H = at2 + 554 H = -0.5(9.8)t2 + 554 H = -0.5(9.8)t2 + 554 H = -4.9t2 + 554

Therefore, H = -4.9t2 + 554 is an equation in standard form that models this relationship.

Example #1 cont’d

Recall, we’re looking for the length of time that the jumper is in free fall.

The parachute opened at 300m, so substitute 300 for H and then find t.

H = -4.9t2 + 554 300 = -4.9t2 + 554 300 -554 = -4.9t2 + 554 -554 -254 = -4.9t2

t2 = 51.84 = t = 7.2

Example #1 cont’d

So the jumper is in free fall for about 7.2 seconds.

Reflect on Example 1 We used 0 for the t-

coordinate because we started counting when he jumped out of the plane

We were only considering the right-half of this graph◦ t 0

A coffee shop sells a special blend of coffee for $2.60 per mug. The shop sells about 200 mugs per day. Customer surveys show that for every $0.05 decrease in the price, the shop will sell 10 more mugs per day.

a) Determine the maximum daily revenue from coffee sales and the price per mug for this revenue.

b) Write an equation in both standard form and vertex form to model this problem. Then sketch the graph.

Example #2

Let x represent the number of $0.05 decreases in price

Revenue = (price)(mugs sold) Price: 2.60 – 0.05x Mugs Sold: 200 + 10x Revenue = (2.60 – 0.05x)(200 + 10x) We want to find the maximum revenue so

we need to vertex We can start by finding the zeros, and then

finding the middle of the zeros to determine the equation of the axis of symmetry

Example #2 cont’d

We find the zeros (x-intercept) by setting y=0

Revenue = (2.60 – 0.05x)(200 + 10x) 0 = (2.60 – 0.05x)(200 + 10x) So either 2.60 – 0.05x = 0 or 200 + 10x = 0 If 2.60 – 0.05x = 0, x = 52 If 200 + 10x = 0, x = -20 The vertex is in the middle of the zeros, so:

Example #2 cont’d

We know the vertex is at x = 16, the r value is: Revenue = (2.60 – 0.05x)(200 + 10x)r = (2.60 – 0.05(16))(200 + 10(16))r = (1.80)(360)r = 648a) Therefore, the maximum daily revenue is $648.The price per mug to maximize revenue is:2.60 – 0.05x at x = 16, so:2.60 – 0.05(16) = 1.80So the coffee shop should sell each mug of coffee for $1.80 to achieve the maximum revenue of $648.

Example #2 cont’d

b) Write an equation in both standard form and vertex form to model this problem. Then sketch the graph.Here, the vertex is (16, 648) = (h, k), so we can write:

r = a(x – 16)2 + 648 We know that when x = 0, r = (2.60 – 0.05x)(200 +

10x), so: r = (2.60)(200) = 520, so a point is (0, 520) – use to find

“a” r = a(x – 16)2 + 648 520 = a(0 – 16)2 + 648 520 = a(– 16)2 + 648 -128 = 256a -0.5 = a So, the equation in vertex form is r = -0.5(x – 16)2 + 648

Example #2 cont’d

To get standard form, expand the equation in vertex form. r = -0.5(x – 16)2 + 648 r = -0.5(x2 – 32x + 256) + 648 r = -0.5x2 + 16x - 128 + 648 r = -0.5x2 + 16x + 520 is the equation in standard form.

We can sketch the graph because we know the vertex: (16, 648), x-intercepts (x = 52 & x = -20) and the y-intercept (0, 520):

Example #2 cont’d

All quadratic relations can be expressed in vertex form and standard form

Quadratic relations that have zeros can also be expressed in factored form

For any parabola, the value of a is the same in all three forms of the equation of the quadratic relation

In Summary…