Math Test 2 Flashcards
Chain Rule
d/dxf(g(x)) = f’(g(x))g’(x) = df/dgdg/dx
Differentiate f(x)=sin(3x)
cos(3x)(3)
Differentiate f(x)=(x^3-2x+1)^6
6(x3-2x+1)5(3x2-2)
Implicit Differentiation
- Taking derivative of both sides using Chain rule
- Solve for dy/dx
Differentiate x^2+y^2=1, then find x where y’=1 using Algebra & Diagram
dy/dx=-2x/2y
x=+/- 1/sqrt(2)
Logarithmic Differentiation
Applying the natural logarithm to both sides of equation to make it easier to differentiate, especially in f(x)^g(x), combining Chain Rule+Implicit Differentiation+Logarithmic Rules
- Take log of both sides, simplifying using log rules
- Take d/dx of both sides (implicit differentiation+chain rule)
- Isolate for y’(x)
Differentiate x^a using logarithmic differentiation
f(x)= x^a
logf(x) = logx^a
1/f(x) * f’(x) = alogx
1/(x^a) * f’(x)=a/x
f’(x) = a/x(x^a)
f’(x)= ax^(a-1)
How to solve a related rate question?
- Draw & label a picture + Write down what is known
- Write down what you wish to find
- Write down an equation that relates the thing whose rate y (reduce to 1 variable)
- Differentiate & Solve
- Check if answer makes sense
Rate of Change of Area = L’W + LW’
Rate of Change of Focal Length of Lense
Rate of Change of radius of expanding circle
Rate of change of water level entering inverted cone
Rate of Change of angle of a moving person across a room
Wall = 5.9m
Rate of Change of Walking = 1/4m/s
x = opposite/walking distance
tangent = angle
At tangent=pi/4, what is the rate of change of tangent?
Draw label diagram
Find dtangent/dt or rate of change of tangent or f’(pi/4) f(x)=tangent
t * tan(tangent)=tx/L –> differentiate ^ solve
(1215.6)1.4
An inverted cylindrical cone, 28 m deep and 14 m in diameter across the top, is being filled with water at a constant rate of 13 m3/min. At what rate is the water rising in the tank when the depth of the water is 1m, 10m, 27m
208/pi(h)^2
Draw sin(x)=arcsin(y) & sin(y)=arcsin(x) at range[-pi/2,pi/2]<->domain[-1,1]
See Diagram
Draw cos(x)=arccos(y) & sin(y)=arccos(x) at range[-0,pi]<->domain[-1,1]
See diagram
Differentiate arcsin(x) in terms of x
d/dxsin(y)=dx/dx
cos(y)dy/dx=1
Triangle + SOH CAH TOA
dydx=11-x2
Differentiate arccos(x) in terms of x
d/dxcos(y)dy/dx=dx/dx
dy/dx=-1/sin(y)
Triangle + SOH CAH TOA
dy/dx=-1/sqrt(1-x2)
Differentiate arccos(e^2x) in terms of x
2e^(2x)/sqrt(1-e^4x)
Draw tan(x)=arctan(y) & tan(y)=arctan(x) at range[-pi/2,pi/2]<->domain[-infinity,infinity]
See diagram
Differentiate arctan(x) in terms of x
d/dxtan(y)=dx/dx
sec^2(y)dy/dx=1
dy/dx=1/sec2(y)
dy/dx=cos^2(y)
Triangle + SOH CAH TOA
dy/dx=1/1+x2
How to determine concavity?
If f(x)’‘>0 = slope is increasingly increasing = concave up = f(x) lies above its tangent lines on that interval
Diagram
If f(x)’‘<0 = slope is increasingly decreasing = concave down = f(x) lies below its tangent lines on that interval
Diagram
Degree 0 & 1(linear) approximation around x=0? Diagram?
T0(x) be the degree 0 approximation of f(x) around x=0: T0(x)=f(0)=1
Let T1(x)be the linear approximation to f(x) about x=0: T1(x)=f(0)+f’(0)(x-a)=1+1(x-0) = 1+x
The higher the degree of approximation, the better the approximation
Degree n Approximation to f(x) at x=a is ?
Tn(x)=f(a)+f’(a)(x-a)+f’‘(a)2(x-a)2+f’’‘(a)3!(x-a)3+…+f(n)(a)n!(x-a)n
Taylor vs Maclaurin Polynomial
Taylor x=a
Maclaurin x=a=0
Proof of Degree nth Approximation
Find c0,c1,c2,c3,…,cn, such that the kth derivatives match: T(k)n(x)=f(k)(a) for k=0,1,2,3,…,n
Proof of Degree 3 e^x Approximation
Let T2(x)=c0+c1(x-0)+c2(x-0)2=c0+c1x+c2x2 be the degree 2 approximation to f(x)=ex about x=0
Find c0, c1 and c2 such that
T2(0)=f(0)=1 → c0+c1(0)+c2(0)2=c0=1
T’2(0)=f’(0)=1 → c1+2c2(0)=c1=1
T’‘2(0)=f’‘(0)=1 → 2c2=1, c2=1/2
Therefore: T2(x)=c0+c1x+c2x2 = 1+1x+1/2x
cos(x) nth Degree Maclaurin Polynomial + Show Work
Tn(x)=1-x2/2!+x4/4!+…
sin(x) nth Degree Maclaurin Polynomial:
Tn(x)=x-x3/3!+x5/5!+…
Find degree 3 MP of f(x)=4xe^2x:
expand: T3(x)=4x[1+(2x)+(2x)2/2!+(2x)3/3!]
Derive f(x)=a1-x general formula of nth degree Maclaurin Polynomial
Tn(x)=a+ax+ax2+ax3+…+ax^n
Find limit of the partial sums Sn as n–>infinity
Sn(x)=a+ax+ax2+…+axn-1
xSn(x)=ax+ax2+ax3+…+axn-1+axn
Sn(x)-xSn(x)=a-axn
Sn=a(1-x^n)/(1-x)
lim n–>infinity a +ax+ax2+…+axn-1 = a/1-x if absolute x < 1 so x^n is negligeble
Make statement about 5-10/3+20/9-40/27+….
lim n–>infinity (5+5(-2/3)+5(-2/3)^2+5(-2/3)^3… )
=5/(1+2/3)=3
Find the sixth degree Taylor Polynomial T6(x) about x=0 of (e^3x^2)/(1+5^x^3)
T6(x)= 1 + 3x^2-5x^3+ 0x^3 + 9/2x^4 - 15x^5 +59/2x^6
A water tank is in the shape of an inverted right circular cone with top diameter 6 metres and depth 8 metres. Suppose that the tank is leaking water. Note the volume of a right circular cone is V = 1/3πr2h, where r is the radius and h is the height of the cone.
a) Suppose that when the water is 4 metres deep, the water leaks out of the tank at a rate of 3 m^3//hr. What is the rate of change of the depth of the water at this point in time?
b) Suppose now that the water from the conical tank is leaking directly into another tank, which is in the shape of a cube, and has base area A m2. How fast is the depth of the water in the cubic tank increasing atthe same moment in time described in part (a)?
a) h’ = 4/3pi m/hr
b) h’c = 3A m/hr
Consider a container filled with some fixed amount of ideal
gas. Suppose that the container is perfectly isolating, the thickness of its wall is negligible and it is capable of expanding or shrinking while staying perfectly spherical (you may think of an ideal balloon). The pressure p, the volume V and the temperature T of the gas satisfy the ideal gas law pV = nRT, where n is the amount of the gas enclosed in moles and R is a universal constant.
You may use that the volume of a solid ball of radius r is V = 4/3πr3 and the surface area of a sphere of radius r is A = 4πr2
(a) Suppose that we heat up the gas at a constant rate of 3 K/s while keeping the pressure at a constant level 3/4 Pa. What is the rate of change of the radius of the container, when the surface area of the container is 0.4 m2? Your answer should be in terms of n and R.
b) Now suppose that we keep the temperature constant while we are expanding the container in such a way that its radius is increasing at a constant rate of 0.1 m/s . What is d/dt log(p) when the radius of the container is 0.6 m?
dr/dt = 10nR.
d/dt log(p) = − 0.3
0.6 = −1/2
Consider an isosceles triangle of base 10 centimetres and
sides x centimetres. Suppose x is increasing at a rate of 4 centimetres per second.
a) What is the rate of change of the height of the triangle when x = 13 centimetres?
b) What is the rate of change of the area of the triangle when x = 13 centimetres?
a) h’ = 13/3
b) A’= 65/3
