Rates of Change Question. How to do this?

The equation of the path of a bullet fired into the air is y = −20x(x − 20), where x and y are displacements in metres horizontally and vertically from the origin. The bullet is moving horizontally at a constant rate of 1 m/s.

(a) Find the rate at which the bullet is rising:
(i) when x=8, , (ii) when y=1500.

(b) Find the height when the bullet is:
(i) rising at 30 m/s, (ii) falling at 70 m/s.

(c) Use the gradient function dy/dx to find the angle of flight when the bullet is rising at 10 m/s.

(d) How high does the bullet go, and how far away does it land?

1 Answer
Jun 21, 2018

Kindly go through the Explanation Section.

Explanation:

Given that, #y=-20x(x-20)=-20x^2+400x...(star)#.

Here, #x# & #y# resp. denote the horizontal & vertical

displacements from the origin of the bullet.

So, #dx/dt, &, dy/dt# are the resp. rates at which the bullet is

moving horizontally & rising.

#"Diff.ing "(star)" w.r.t. "t," dy/dt=-40xdx/dt+400dx/dt#.

As, #dx/dt=1"m/s, "dy/dt=-40x+400....................(star_1)#.

Part (a)(i)

# "When "x=8, "the reqd. rising rate, i.e., "[dy/dt]_(x=8)#

#=-40(8)+400#,

#=80"m/s"#.

(a)(ii):

#y=1500. :. (star) rArr -20x^2+400x=1500#.

#:. x^2-20x+75=0#.

#:. (x-15)(x-5)=0," giving, "x=15, or, 5#.

#:. (star_1)" gives "[dy/dt]_(x=15)=200"m/s, or, "#

#[dy/dt]_(x=5)=200"m/s"#.

Part (b)(i):

The bullet is rising at #30"m/s. ":. dy/dt=30#.

#:. -40x+400=30 rArr x=(400-30)/40=37/4#.

At this #x#, the reqd. height #y# of the bullet is given by,

#y=-20*37/4(37/4-20)=5*37*43/4=7955/4=1988.75"m"#.

(b)(ii):

The height #y# when the bullet is falling at #70"m/s"# can

similarly be worked out by taking #dy/dt=color(red)(-70)#.

Part (c):

We require #dy/dx=-40x+400," when "dy/dt=10#.

Now, #dy/dt=10rArr -40x+400=10 rArr x=39/4#.

So, #[dy/dx]_(x=39/4)=-40*39/4+400=10#.

But, we know that #dy/dx# gives the slope of tangent.

So, if the angle of flight is #psi#, then, we have,

#tanpsi=[dy/dx]_(x=39/4)=10#.

#:. psi=arctan 10#.

Part (d):

For the maximum height #y_max# of the bullet, we must have,

#dy/dx=0, and, (d^2y)/dx^2 lt 0#.

#:.-40x+400=0:.x=10,"&, [(d^2y)/dx^2]_(x=10) lt 0#.

#:. y_max=-20*10(10-20)=2000"m#.