The most common time that we use derivatives is when we are finding the rate of change with respect to time. +Example 72 km/h > top speed 1.5 km > how long it will take v = 72 km/h d = 1.5 km t = ? 72 km/h = 1.5 km/t multiply each side by t. 72t = 1.5 t = 1.5 km/72 km/hr = .021 hr .021 hr X 60 min/1 hr = 1.26 min
Derivatives can also be used when dealing with acceleration. a = vf - vi/t +Example d = 10.0 > 10X15 = 150 m t = 60.0 s v = 150 m/60.0 s = 2.5 m/s t = .25 s a = 2.5 m/s - 2.5 m/s^2/.25 = 0/.25 = 0 m/s^2
One other way to use derivatives when dealing with velocity. v = ds/dt +Example: Flying 1.73 km downward in 25 s. v = 1.73 km/25s = .0692 km/s
There are many things that you use to find derivatives like product rule, quotient rule, when you want to find the slope of a normal line, when you are trying to find the rate of change, to find a slope of a horizontal tangent line, and when you want to find the equation of a tangent line. To find the derivative when using the product rule: (recopy 1st)(derivative 2nd)+(recopy 2nd)(derivative 1st). (3x-2x^2)(5+4x) (3x-2x^2)(4)+(5+4x)(3-4x) 12x-8x^2+15-20x+12x-16x^2 = -24x^2+4x+15
To find the quotient rule: (bottom)(derivative top)-[(top)(derivative bottom)]/(bottom)^2 5x-2/x^2+1 (x^2+1)(5)-[(5x-2)(2x)]/(x^2+1)^2 5x^2+5-[10x^2-4x]/(x^2+1)^2 5x^2+5-10x^2+4x/(x^2+1)^2 = -5x^2+4x+5/(x^2+1)^2
To find the slope of the line tangent to the graph of the function f(x) = 5-x^2 at the point (2, 1). You take the derivative and then plug the x value in. First you take the derivative of 5-x^2 which is -2x. then you plug the x value in for x. -2(2) = -4 that is what your slope is equal to.
Three example that use derivatives are doing a problems involving Roolle's theorem, the first derivative test, the second derivative test. here are example problems using each concept:
Roolle's Theorem:
let f(x)=x^4-2x^2. find all values c in the interval (-2,2) such that f^1(c)=0.
Derivatives are used to find a number of things, but I think the three most obvious are the slope of a tangent line, a rate of change, and can be used in finding minimums and maximums.
ex. slope of tangent line:
2x^2 - 3x slope of line tangent to this equation is 4x - 3 because that is the derivative.
Derivatives are used to find the slope of a tangent line, a rate of change, and finding minimums and maximums....Oh and just throwin this out there, I just maxed 295 on bench press!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
okay: slope of a tangent line:
3 x^2 - 4 x + 2 The derivative of this would be : 6 x - 4 This is the slope of a tangent line.
Rate of Change :
I don't really remember this too well and I forgot my binder at school so I'm using Matt's example...
You take the derivative as like the first step and then if I'm not mistaken you end up taking it twice further down the line, I wish I had my binder because I feel pretty stupid right now....butttttt yeah it's definitely needed in finding max's and min's.
they are used to find the slope of a tangent line, you can use roolles therom and use product rule, quotient rule, ex. slope of tangent line: 2x^2 - 3x slope of line tangent to this equation is 4x - 3 because that is the derivative.
Roolle's Theorem:
let f(x)=x^4-2x^2. find all values c in the interval (-2,2) such that f^1(c)=0.
1. yes
2. yes
3. f(-2)=8 f(2)=8
4. 4x^3-4x=0 4x(x^2-1)=0
x=0,+/-1
To find the derivative when using the product rule: (recopy 1st)(derivative 2nd)+(recopy 2nd)(derivative 1st). (3x-2x^2)(5+4x) (3x-2x^2)(4)+(5+4x)(3-4x) 12x-8x^2+15-20x+12x-16x^2 = -24x^2+4x+15
The most common time that we use derivatives is when we are finding the rate of change with respect to time.
ReplyDelete+Example
72 km/h > top speed
1.5 km > how long it will take
v = 72 km/h
d = 1.5 km
t = ?
72 km/h = 1.5 km/t
multiply each side by t.
72t = 1.5
t = 1.5 km/72 km/hr
= .021 hr
.021 hr X 60 min/1 hr
= 1.26 min
Derivatives can also be used when dealing with acceleration. a = vf - vi/t
+Example
d = 10.0 > 10X15 = 150 m
t = 60.0 s
v = 150 m/60.0 s
= 2.5 m/s
t = .25 s
a = 2.5 m/s - 2.5 m/s^2/.25
= 0/.25
= 0 m/s^2
One other way to use derivatives when dealing with velocity. v = ds/dt
+Example:
Flying 1.73 km downward in 25 s.
v = 1.73 km/25s
= .0692 km/s
There are many things that you use to find derivatives like product rule, quotient rule, when you want to find the slope of a normal line, when you are trying to find the rate of change, to find a slope of a horizontal tangent line, and when you want to find the equation of a tangent line.
ReplyDeleteTo find the derivative when using the product rule: (recopy 1st)(derivative 2nd)+(recopy 2nd)(derivative 1st).
(3x-2x^2)(5+4x)
(3x-2x^2)(4)+(5+4x)(3-4x)
12x-8x^2+15-20x+12x-16x^2
= -24x^2+4x+15
To find the quotient rule: (bottom)(derivative top)-[(top)(derivative bottom)]/(bottom)^2
5x-2/x^2+1
(x^2+1)(5)-[(5x-2)(2x)]/(x^2+1)^2
5x^2+5-[10x^2-4x]/(x^2+1)^2
5x^2+5-10x^2+4x/(x^2+1)^2
= -5x^2+4x+5/(x^2+1)^2
To find the slope of the line tangent to the graph of the function f(x) = 5-x^2 at the point (2, 1). You take the derivative and then plug the x value in.
First you take the derivative of 5-x^2 which is -2x. then you plug the x value in for x.
-2(2) = -4 that is what your slope is equal to.
Three example that use derivatives are
ReplyDeletedoing a problems involving Roolle's
theorem, the first derivative test,
the second derivative test.
here are example problems using each
concept:
Roolle's Theorem:
let f(x)=x^4-2x^2. find all values
c in the interval (-2,2) such that
f^1(c)=0.
1. yes
2. yes
3. f(-2)=8
f(2)=8
4. 4x^3-4x=0
4x(x^2-1)=0
x=0,+/-1
First Derivative Test:
find the relative extrema of
f(x)=(x^2-4)^2/3
1. 2/3(x^2)^-1/3 x 2x=0
4x/3(x^2-4)^1/3=0
x=0
2. x^2-4=0
x=+/-2
3. (-infinity,-2)u(-2,0)u(0,2)u(2,infinity)
4 f(-3)=-ve-dec
f(-1)=+ve-inc
f(1)=-ve-dec
f(3)=+ve-inc
max= x=0
min= x=-2,2
Second Derivative Test:
f(x)= 6/x^2+3
determine the open intervals where
the function is concave up or
concave down.
1. f(x)= -12x/(x^2+3)^2
f(x)=36(x^2-1)/(x^2+3)^2=0
x=+/-1
2. x^2+3=0
x^2=3
x=square root of 3i
3.(-infinity,-1)u(-1,1)u(1,infinity)
4. f(-2)=+ f(0)=- f(2)=+
concave up concave down concave up
x=-1 is a point of inflection
x=1 is a point of inflection
concave up: (-infinity, -1)u(1,infinity)
concave down: (-1,1)
Derivatives are used to find a number of things, but I think the three most obvious are the slope of a tangent line, a rate of change, and can be used in finding minimums and maximums.
ReplyDeleteex. slope of tangent line:
2x^2 - 3x
slope of line tangent to this equation is 4x - 3 because that is the derivative.
ex. rate of change:
dv/dt = ? r = 4 dr/dt = 2
dv/dt = 4/3pi3r^2dr/dt
dy/dt = 384pi
--by taking the derivative of the formula for volume then pluggin in, you show the volumes rate of change.
when finding maxs an mins taking the derivative is also a very important step
Derivatives are used to find the slope of a tangent line, a rate of change, and finding minimums and maximums....Oh and just throwin this out there, I just maxed 295 on bench press!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ReplyDeleteokay: slope of a tangent line:
3 x^2 - 4 x + 2
The derivative of this would be :
6 x - 4
This is the slope of a tangent line.
Rate of Change :
I don't really remember this too well and I forgot my binder at school so I'm using Matt's example...
dv/dt = ? r = 4 dr/dt = 2
dv/dt = 4/3pi3r^2dr/dt
dy/dt = 384pi
Finding max's and min's :
You take the derivative as like the first step and then if I'm not mistaken you end up taking it twice further down the line, I wish I had my binder because I feel pretty stupid right now....butttttt yeah it's definitely needed in finding max's and min's.
they are used to find the slope of a tangent line, you can use roolles therom and use
ReplyDeleteproduct rule, quotient rule,
ex. slope of tangent line:
2x^2 - 3x
slope of line tangent to this equation is 4x - 3 because that is the derivative.
Roolle's Theorem:
let f(x)=x^4-2x^2. find all values
c in the interval (-2,2) such that
f^1(c)=0.
1. yes
2. yes
3. f(-2)=8
f(2)=8
4. 4x^3-4x=0
4x(x^2-1)=0
x=0,+/-1
To find the derivative when using the product rule: (recopy 1st)(derivative 2nd)+(recopy 2nd)(derivative 1st).
(3x-2x^2)(5+4x)
(3x-2x^2)(4)+(5+4x)(3-4x)
12x-8x^2+15-20x+12x-16x^2
= -24x^2+4x+15