Rates of Change Question. How to do this?

Question

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?

Charles Arocha · Accepted Answer

Kindly go through the Explanation Section.

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#.

Cameron Alexander · Answer

undefined To answer questions about rates of change, you typically use calculus, specifically differentiation. Here are the general steps:

1. **Understand the Problem**: Identify what quantity is changing (dependent variable) and what it is changing with respect to (independent variable).

2. **Set Up the Function**: Write down the function that describes the relationship between the variables. For example, if you have distance changing over time, you might have a function $ D(t) $ where $ D $ is distance and $ t $ is time.

3. **Differentiate the Function**: Take the derivative of the function with respect to the independent variable. This gives you the rate of change of the dependent variable with respect to the independent variable.

4. **Interpret the Result**: Once you have the derivative, it will be in terms of the independent variable. If it's a physical quantity like distance, it might be in units like meters per second. If it's a business problem, it might be in terms of currency per unit of time, etc.

5. **Solve the Problem**: Use the derivative to answer specific questions about rates of change. For example, finding when the rate of change is zero (critical points), finding maximum or minimum rates of change, etc.

If you have a specific question or problem related to rates of change, please provide the details, and I can guide you through the steps.