# Tutoring math, you encounter the differences between calculators even from the same manufacturer. Now, the tutor shows how to convert from rectangular to polar coordinates using the Casio fx-991ES PLUS C.

In my post from May 18 I show how to convert from rectangular to polar coordinates using the Casio fx-260solar. Here’s how to do the same with the Casio fx-991ES PLUS C:

Example: Convert (-80,-56) to polar coordinates using the Casio fx-991ES PLUS C.

Solution:

Key in SHIFT + -80 SHIFT ) -56 ) =

Across the bottom, the length (r) will show first; arrow over to see θ.

HTH:)

Source:

Casio fx-991ES PLUS C User’s Guide.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring calculus, you cover polar coordinates. The tutor shows how to convert from rectangular to polar coordinates on the Casio fx-260solar.

Example:

Convert the coordinates (-56,12) to polar with the Casio fx-260solar.

Solution:

1. Key in 56 +/- SHIFT + 12 =
2. Hopefully you get 57.27, which is the length.
3. Now key in SHIFT, then the left bracket key.
4. Hopefully you get 167.9, which is the angle in degrees.

HTH:)

Source:

Casio fx-260solar manual

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor uses l’Hôpital’s rule to find a limit of form ∞/∞.

l’Hôpital’s rule states that the limit of a quotient of form ∞/∞ or 0/0 can be found as follows:

lim (f(x)/g(x)) = lim (f'(x)/g'(x))

In this case [noting the square root of x is x0.5]:

limx→∞(lnx/x0.5) = (by l’Hôpital) limx→∞((x-1)/(0.5x-0.5))

which becomes

limx→∞2x-0.5 or limx→∞2/x0.5 = 0

By that reasoning, the reciprocal limit, limx→∞(x0.5/lnx), should not exist.

Source:

Larson, Roland E. and Robert P. Hostetler. Calculus, 3rd ed. Toronto: DC Heath, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor shows the derivative of arcsin, the inverse of sin.

In yesterday’s post I explained the formula for the derivative of an inverse function

(m-1(x))’ = 1/m'(m-1(x))

Today, I’ll use it to find the derivative of “inverse sin(x)”, aka sin-1(x), aka arcsin(x).

arcsin'(x) = 1/cos(arcsin(x))

Now, behold:

We see, in the illustration, that arcsin(x) = θ. The formula becomes

arcsin'(x) = 1/cos(θ)

Once again, from the illustration: cos(θ) =

So we have

arcsin'(x) = 1/

Source:

Larson, Roland E. and Robert P. Hostetler. Calculus, 3rd ed. Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor shows the development of a formula for the derivative of an inverse.

Let’s imagine m(x) is a function with inverse m-1(x). Then

m(m-1(x)) = x

By implicit differentiation,

[m(m-1(x))]’ = 1

By the chain rule,

[m(m-1(x))]’ = m'(m-1(x))*(m-1(x))’

Therefore,

m'(m-1(x))*(m-1(x))’ = 1

Dividing both sides by m'(m-1(x)) yields

(m-1(x))’ = 1/m'(m-1(x))

In a coming post I’ll show an example of using this formula to find the derivative of a specific inverse function.

HTH:)

Source:

Larson, Roland E. and Robert P. Hostetler. Calculus, 3rd ed. Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor shows the example ∫dx/(x2+6)

∫dx/(x2+1) = arctanx + C

The related integral

∫dx/(x2+6)

must be put in the form, as follows:

∫dx/(x2+6) = ∫dx/(6(x2/6+1)) = 1/6 ∫dx/(x2/6 + 1)

=1/6 ∫dx/((x/√6)2+1) = (√6)/6∫(dx(1/√6))/((x/√6)2 + 1)

Next it becomes

1/√6∫(dx(1/√6))/((x/√6)2 + 1)

which can be integrated:

1/√6 ∫(dx(1/√6))/((x/√6)2 + 1)=(1/√6)arctan(x/√6) + C

HTH:)

Source:

Larson, Roland E. and Robert P. Hostetler. Calculus, 3rd ed. Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor looks at forming a Taylor polynomial with the example of square root 31.

A transcendental function is one there is no operation for. Rather, it’s represented by a series of expressions. Square root and sin are two examples.

The Taylor polynomial for a function is defined as

P(x)=f(c) + f'(c)(x-c) + f”(c)(x-c)2/2! + f”'(c)(x-c)3/3! + ….

For presentation purposes, we note that square root c = c0.5. In general,

Following the form of the Taylor polynomial gives, for square root,

P(x) = (c)0.5 + 0.5c-0.5(x-c) -0.25c-1.5(x-c)2/2! + 0.375c-2.5(x-c)3/3! – 0.9375c-3.5(x-c)4/4! + ….

The meaning of c

In the Taylor polynomial above, c is an “anchor value” at which you already know the output. Preferably it’s the closest value [to the one being evaluated] for which the exact answer is known.

Example: Evaluate square root 31 using a Taylor polynomial.

Solution: closest to 31 is 36, so c=36. Then

P(31) = 360.5 + 0.5(36)-0.5(31-36) – 0.25(36)-1.5(31-36)2/2! + 0.375(36)-2.5(31-36)3/3! – 0.9375(36)-3.5(31-36)4/4! + ….

which becomes

6 + (0.5/6)(-5) – (0.25/216)(-5)2/2! + (0.375/7776)(-5)3/3! – (0.9375/279936)(-5)4/4! + ….

and then

6 – 0.416666667 – 0.014467592 – 0.00100469393 – 0.00008721301476

=5.567773835

According to the calculator,

310.5 = 5.567765363

The difference between the values is 0.00000947215. Perhaps each term in the series gives an additional decimal place of accuracy.

HTH:)

Source:

Larson, Roland E. and Robert P. Hostetler. Calculus, 3rd ed. Toronto: DC Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor explains concavity and point of inflection with an example.

Concavity refers to an aspect of graph shape. My first-year calculus professor explained it this way: concave upward will collect rain, while concave downward will shed rain. Numerically, when the second derivative is positive, the graph is concave upward. When the second derivative is negative, the graph is concave downward.

In the graph above, the section from P to Q is concave downward; from Q to R is concave upward.

A point where concavity changes from negative to positive (or positive to negative) is called a point of inflection. In the graph above, Points Q and R are inflection points.

At a point of inflection, the second derivative is either 0 or undefined. However, f”=0 doesn’t guarantee a point of inflection; you still have to check either side (if you can’t see the graph).

The graph above, y=sinx, has second derivative -sinx. At point Q (where x=Π), -sin(Π)=0. Just to the left, -sin3=-0.1411. (Recall Π=3.14159….) Past Π, -sin3.3=0.1577. The sign change across Π confirms the inflection point at (Π,0).

Source:

Larson, Roland and Robert Hostetler. Calculus, part one. Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor shows how to remind yourself that the derivative of tanx is sec2x.

Let’s imagine you don’t recall that (tanx)’=sec2x. Here’s how to reconstruct it:

1. Recall that tanx=sinx/cosx
2. Take the derivative of sinx/cosx using the quotient rule:(u/v)’ = (vu’-uv’)/v2
3. In this case, (sinx/cosx)’ = [cosx(cosx) – sinx(-sinx)]/cos2x
4. Recall that cos2x +sin2x = 1.
5. Simplifying we arrive at (sinx/cosx)’ = 1/cos2x = sec2x

Therefore, (tanx)’=(sinx/cosx)’=sec2x.

Source:

Larson, Roland and Robert Hostetler. Calculus, part one. Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor explores how to detect and solve exact differential equations with a very simple example.

When a differential equation of the form

P(x,y) +Q(x,y)y’ = 0

results from the implicit differentiation of an original equation F(x,y)=c, the equation

P(x,y) +Q(x,y)y’ = 0

is said to be an exact differential equation.

The way to tell is that, for an exact differential equation,

Py(x,y) = Qx(x,y)

Example: Solve the differential equation

siny +1 +xy’cosy +2y’ = 0

First, we render it to P(x,y) +Q(x,y)y’ = 0:

siny + 1 +(xcosy +2)y’ = 0

P(x,y) = siny+1; Q(x,y) = xcosy + 2

Now we take the “other derivative” of each one:

Py(x,y) = cosy

Qx(x,y) = cosy

Py(x,y) = Qx(x,y): the equation is exact. Therefore,

siny +1 +xy’cosy +2y’ = 0

is the implicit derivative of some equation F(x,y) = c, which we need to find. Furthermore,

siny + 1 = Fx

We integrate siny + 1 with respect to x:

∫siny +1 = xsiny + x + g(y)

where g(y) is a function only of y that was lost in the original derivative by x.

g'(y) should be recognizable in Q(x,y): specifically, it’s 2⇒ g(y)=2y

Therefore, the solution F(x)=c is

xsiny +x +2y=c

Note that its implicit derivative is

siny + 1 + x(cosy)y’ +2y’ = 0

which matches the original differential equation posed.

Source:

Boyce, William and Richard DiPrima. Elementary Differential Equations and Boundary Value Problems. Toronto: John Wiley & Sons, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.