Lesson: _____ Section 6.3,5 Intro to Differential Equations & The Equations of Motion An equation containing derivatives is called a “differential equation.” Ex. A car is driving in a straight line at a constant velocity of 50 mph. If s is the distance of the car from a fixed point, and t is time in hours, then This is called “rectilinear motion” (straight line) This is called a “differential equation” for function s. This says the derivative of s is 50. 𝑑𝑠 𝑑𝑡 =50 50𝑑𝑡 The “solution” to the diff. eq. is function s, which is the antiderivative of 50 𝑠= =50𝑡+𝑐 This is a general solution. To make it specific, we would need to know an initial value. In this case, when 𝑡=0 we know that s = c, therefore c represents the initial distance from the reference point.
This diff. eq. represents the acceleration. Why is it negative? Ex. An apple is dropped from a 100 ft building. Find position and velocity as a function of time. When does it hit the ground? How fast is is going when it hits? Acceleration due to gravity 𝑔=9.8 𝑚 𝑠 2 𝑜𝑟 32 𝑓𝑡 𝑠 2 If distance is in feet, and time is in seconds, then… 𝑑𝑣 𝑑𝑡 =−32 This diff. eq. represents the acceleration. Why is it negative? 𝑣=−32𝑡+c Since 𝑣=𝑐 when 𝑡=0 then c must represent our initial velocity. Let’s call it 𝑣 0 . 𝑣=−32𝑡+ 𝑣 0 Since the apple was dropped, 𝒗 𝟎 = 𝑣=−32𝑡 𝟎 𝑑𝑠 𝑑𝑡 Now, 𝑣= =−32𝑡 Let’s rewrite this function for velocity as a differential equation for position 𝑠=−16 𝑡 2 +𝑘 𝑠=−16 𝑡 2 + 𝑠 0 Since 𝑠=𝑘 when 𝑡=0 then k must represent our initial position. Let’s call it 𝑠 0 . 𝑠=−16 𝑡 2 +100 We know that 𝒔 𝟎 =𝟏𝟎𝟎
When does it hit the ground? How fast is is going when it hits? Think about it. What does it mean in terms of position when the apple hits the ground? 𝒔=𝟎 Since 𝑠=−16 𝑡 2 +100 We want to solve the equation: 0=−16 𝑡 2 +100 16 𝑡 2 =100 𝑡 2 = 100 16 𝑡=± 100 16 =± 10 4 =± 2.5 The apple hits after 2.5 seconds This is asking for 𝑣(2.5) 𝑣=−32𝑡 𝑣 2.5 =−32 2.5 =−80 𝑓𝑡 𝑠 Should this be negative? Yes! Our position is decreasing which means a negative rate of change. See also p. 295 example 2. Note initial velocity is given. How do we find the “highest point” using the derivative? Take a minute to READ 6.5 (history of the equations of motion)