math 19a: modeling and differential equations for the life sciences calculus review danny kramer...
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Math 19a: Modeling and Differential Equations for the Life Sciences
Calculus Review
Danny KramerFall 2013
Point Slope Concept
π β² (π₯ )= limβπ₯β 0
π (π₯+β π₯ )β π (π₯)β π₯
πππ₯
π (π₯ )=π ππππππ‘ πππ¦ πππππ‘ π₯
Think of , but change in y measured over infinitely small change in x
x
y
Derivative Rules
πππ₯
π=0
πππ₯
π₯π=ππ₯πβ1
πππ₯
π ππ π₯=βπ πππ₯
πππ₯
π πππ₯=πππ π₯
πππ₯ln β¨π₯β¨ΒΏ
1π₯
πππ₯
ππ₯=ππ₯
All with respect to dx, ie if youβre using 2x, then put 2x in for x and 2 in front of all derivatives.
Derivatives and Operations
πππ₯
( π +π)= π β²+π β²
πππ₯
( π βπ)= π β²βπ β²
πππ₯
( π βπ)= π β²π+ ππ β²
πππ₯
ΒΏ
πππ‘π :πππ₯
π= π β²
Applications
β’ Positionβ’ Speed/Velocityβ’ Acceleration
π£=βπ₯β π‘
a=β π£β π‘
π£ (π‘ )= πππ‘
π₯=π₯ β² (t)
a (t )= πππ‘
π£= πππ‘ ( πππ‘ π₯)= π2
ππ‘ 2π₯=π₯ β² β² (π‘)
Some Vocabulary
β’ Continuous- no holes or jumps in the graph
β’ Differentiable- continuous graph with a derivative at each pointβ¦no βcuspsβ
β XX
β X X
Area Concept
π·ππππ£ππ‘ππ£πββππ’πππ‘ππ‘π¦β π‘πππ
=πππ‘π
It is area under a curve, but think of it more generally as multiplying a changing rate by the elapsed time over which the rate occurs, giving you the change in quantity that the rate is measuring.
x
y
=
Antiderivative Rules and Operations
Whatβs with the C? Disappears in derivative!
β«ππ₯=ππ₯+πΆβ«πππ π₯=π πππ₯+πΆβ« π πππ₯=βπππ π₯+πΆ
β«( π +π)=β« π +β«π β«( π βπ)=β« π ββ«π
U substitution
Replace to visualize
β«πsin ( π₯)cos (π₯)ππ₯ββ«πuππ’=ππ’+πΆβπ’=sin (π₯)ππ’=cos (π₯)ππ₯
ππ πππ₯+πΆ
Integration by Parts
β«π’ππ£=π’π£ββ«π£ππ’ Opposite of product rule. Test it out!
What Becomes u?LogInverse Trig (the arcs)AlgebraTrigExponential
β«π₯ πβπ₯ππ₯π’=π₯ππ’=ππ₯π£=βπβπ₯ππ£=πβπ₯ππ₯
Taylorβs Formula
π (π₯ )= π (π)+ π β² (π ) (π₯βπ )+ π β² β² (π)2 !
(π₯βπ)2+β¦
π=βπ=0
β π π(π)π !
(π₯βπ )π
Practicing Taylor
π (π₯ )=π₯ πβπ₯
π β² (π₯ )=πβπ₯βπ₯ πβπ₯=πβπ₯(1βx )
π β² β² (π₯ )=βπβπ₯βπβπ₯ (1β x )=πβπ₯(xβ2)
Practicing Taylor
π (1 )=(1 )πβ1=ππ
=
π β² β² (1 )=βπβ1βπβ1 (1β1 )=πβ1 (1β2 )=βππ
π= π (π )+ π β² (π) (π₯βπ )+ π β² β²(π)2 !
(π₯βπ)2 ,π=1
π π₯=ππ
+πβπ2π
(π₯β1)2
Dimensions of Measurement
β’ x(t) , y(t) x(y) / y(x) ?β’ Match up x and y at any given time t.
t
x , yx y
5
10
x
y
5
10
5 10t0 tf
t0
tf
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