MAT 21A Lecture Notes - Lecture 11: University Of Manchester, Marginal Cost, Marginal Revenue
MAT 21A – Lecture #11 – The Derivative as a Rate of Change
• Derivatives tell us two things:
o 1) The slope of the tangent line to the graph at a particular point
o 2) The rate of change (at a particular instant)
• If f(x) is a function,
is the (instantaneous) rate of change of f (output) with
respect to x (input). If f(t) is a function of time,
or f’t is the instantaneous
rate of change of f at a time, t.
• Recall that the average rate of change of f over [a,b] is equal to
.
• If the positio of tie, t, is pt the p’t is the velocity or rate at which the
positio hages. P’’t =
is called acceleration which describes how
fast the velocity is changing.
• Problem: A ball is thrown upward in the air. The height, in feet, of the ball after t
seconds is h(t) = -16t2 + 20t + 4.
o A) Find the height of the ball after 3 seconds.
h(3) = -16(3)2 + 20(3) + 4 = -80 ft
o B) When does the ball hit the ground?
Set h(t) = 0 0 = -16t2 + 20t + 4 0 = -4(4t2 - 5t - 1). Since this is not
factorable, you would need to use the quadratic formula to solve this.
Recall:
. x =
. Pick the positive solution of
1.4254 seconds since ball took off after t = 0.
o C) What is the velocity after 3 seconds?
I other ords, it is askig ou to fid h’3 so h’t = -32t + 20 h’3 = -
32(3) + 20 = -96 + 20 = -76 ft/s
o D) Find the maximum height of the ball.
To find the highest point that the ball reaches, we need to know when
velocity changes from positive to negative. It is the point right before the
ball starts to fall down.
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Document Summary
If f(x) is a function, (cid:3031)(cid:3033)(cid:3031) is the (instantaneous) rate of change of f (output) with respect to x (input). If f(t) is a function of time, (cid:3031)(cid:3033)(cid:3031)(cid:3047) or f"(cid:894)t(cid:895) is the instantaneous: recall that the average rate of change of f over [a,b] is equal to (cid:3033)(cid:4666)(cid:3029)(cid:4667) (cid:3033)(cid:4666)(cid:3028)(cid:4667) (cid:3029) (cid:3028) positio(cid:374) (cid:272)ha(cid:374)ges. P""(cid:894)t(cid:895) = (cid:3031)(cid:3031)(cid:3047)[(cid:1868) (cid:4666)(cid:4667)] is called acceleration which describes how. If the positio(cid:374) of ti(cid:373)e, t, is p(cid:894)t(cid:895) the(cid:374) p"(cid:894)t(cid:895) is the velocity or rate at which the fast the velocity is changing: problem: a ball is thrown upward in the air. Set h(t) = 0 0 = -16t2 + 20t + 4 0 = -4(4t2 - 5t - 1). Since this is not factorable, you would need to use the quadratic formula to solve this. Recall: (cid:1876)= (cid:3029) (cid:3029)2 (cid:2872)(cid:3028)(cid:3030) (cid:2870)(cid:3028) (cid:4666)(cid:2782)+ (cid:2781) (cid:4667) 1. 4254 seconds since ball took off after t = 0. x = (cid:2869)8(cid:4666)(cid:887) (cid:886)(cid:883) .
