Logs & Exponentials & Inverse Trig Functions

Logarithms & Exponentials

Recall!

The number e

Problem : Determine the balance due on $1.00 invested for 1 year at an interest rate of 100% per annum (interesting bank!) and compounded quarterly, monthly, daily, hourly, half-hourly and "minutely". Recall, the general formula for compounding interest is . For our problem, P = 1, t = 1, r = 1 and n varies. The formula therefore reduces to . Lets make a table:

The table suggests that as n gets bigger and bigger, the balance continues to grow but will never get past $2.72! In other words, exists and is about 2.718. The limit is called " e ". The e is to honour a famous mathematician named Leonard Euler. e is an irrational number and is shown below calculated to 60 significant digits.

> evalf(exp(1),60);

In calculus, we will use e for the exponential and logarithmic functions. Why? Because .

Properties of Exponents.

If a and b are any Real numbers then:

1) =

2)

3)

4)

5)

6)

7) n-th roots : = b <=> . Note: If n is even then n-th roots are only defined if a >0.

Simplify the following:

> e1:=(27)^(-2/3):e2:=(1/16)^(3/4):e3:=((3*x^(-3)*2*y^(2/5)))/((3*x)^3*y^(-2)):

> simplify([e1,e2,e3]);

> plot([10^x,exp(1)^x,2^x],x=-5..5,y=0..4,colour=[black,red,magenta]);

> Limit(exp(1)^x,x=-infinity)=limit(exp(1)^x,x=-infinity);

> Limit(exp(1)^x,x=infinity)=limit(exp(1)^x,x=infinity);

> plot([10^(-x),exp(1)^(-x),2^(-x)],x=-5..5,y=0..4,colour=[black,red,magenta]);

Logarithms

Since f(x) = is a 1-1 function for a > 0 and it follows that the exponential function in base a has an inverse, . It is so famous and oft used that it is given a special name: the base a logarithmic function. . It is written .

Definition : If a > 0 then <=> . If then we write , the natural base logarithm .

Defined as above, it follows that f(g(x)) = x = g(f(x)) where f and g are the exponential and logarithmic functions respectively. i.e. theye are defined in such a way as to be inverses of each other. CHECK!

The domain of the log function is (0, ), which is the range of the exponential function.

The range of the log function is R, which is the domain of the exponential function.

e.g. Find the domain of . Plot. Use the graph to find the range.

> f:=t->ln(t^2-1);

1) Domain.

> solve(t^2-1>0);

> plot(t^2-1,t=-2..2,y=-2..2,colour=black);

From both lines above we can see that the domain is precisely ( ,-1) U (1, ) .

> plot (f(t),t,y=-4..4,numpoints=1000,colour=black);

From the graph, it is clear that the range of f is R .

e.g.'s of properties and non-properties.

> Log[2](16)=simplify(log[2](16));Log[10](1/100)=simplify(log[10](1/100));

> a1:=ln(5*7);b1:=ln(5)*ln(7);c1:=ln(5)+ln(7);

> evalf([a1,b1,c1]);

> a2:=ln(5/7);b2:=ln(5)/ln(7);c2:=ln(5)-ln(7);

> evalf([a2,b2,c2]);

> a3:=ln(3^5);b3:=(ln(3))^5;c3:=5*ln(3);

> evalf([a3,b3,c3]);

Properties of the natural logarithm:

N.B. These properties hold for logarithms of any legal base .

1)

2)

3) .

4)

5)

6)

> plot({exp(1)^x,ln(x),x},x=-8..8,y=-8..8,colour=[black,red,magenta],numpoints=1000);

> Limit(ln(x),x=0)=limit(ln(x),x=0);Limit(ln(x),x=infinity)=limit(ln(x),x=infinity);

e.g. Solve (i) ln(x-1)=1-ln(x+2) and (ii) .

i)

> eq1:=ln(x-1)=1-ln(x+2);

> solve(eq1,x);

> evalf(%);

> solve(x^2+x-2-exp(1)=0);

> evalf(%);

The nicest exact solution is Note: the other is not a solution! Why?

ii)

> eq2:=2^(x+3)=5;

> solve(eq2);

> evalf(%);

check:

> 2^(-.67807+3);

Calculus

> restart:

> f:=(h)->(ln(x+h)-ln(x))/h;

> f(h);

> plot({f(1),f(1/2),f(1/10)},x=-4..4,y=-4..4,numpoints=1000,colour=[black,red,magenta]);

The plots suggest that as h gets "small," the limit approaches , which it does!

> Limit(f(h),h=0)=limit(f(h),h=0);

Theorem: .

Proof: See blackboard.

Chain rule: = f'(x)

Theorem: .

Proof: By implicit differentiation. Do it.

Chain Rule: Obvious!

e.g.'s

Differentiate the following expressions:

> e1:=exp(1)^sqrt(x);e2:=x*exp(1)^(2*x);e3:=tan(exp(1)^(3*x-2));

> diff(e1,x);diff(e2,x);diff(e3,x);

> e4:=x*ln(x);e5:=ln(x^2+1);e6:=ln(exp(1)^(x^2)+ln(x));

> diff(e4,x);diff(e5,x);diff(e6,x);

e.g. Find the tangent line to at (-1,0) and (1,0). Plot.

> restart:with(plots):

> curve:=y=ln(x^2+y^2);

> slope:=implicitdiff(curve,y,x);

> subs(x=-1,y=0,slope);

> subs(x=1,y=0,slope);

> line1:=(y-0)=-2*(x-(-1));

> line2:=(y-0)=2*(x-1);

> implicitplot({curve,line1,line2},x=-4..4,y=-4..4,colour=black);

> y:=a^x;

> diff(y,x);

Other Bases

Use implicit differentiation to find formulas for and and their chain rule generalizations where a > 0 and .

> restart:with(plots):

> y1:=y=a^x;

> implicitdiff(y1,y,x);

> y2:=y = log[a](x);

> implicitdiff(y2,y,x);

Theorem : and . Chain rule generalizations: obvious!

> e1:=3^(exp(1)^x*sin(x^2));e2:=log[2](t^4-t^2+1);

> diff(e1,x);diff(e2,t);

Inverse Trigonometric Functions

The Inverse Sine function.

The sine function is so famous and so important that it demands an inverse. However, if you look at the graph of y = sin(x) you see right away that it is not 1-1 and so has no inverse!

> plot(sin(x),x=-20..20,colour=black);

You can see right away that sin(a) = sin(b) does not imply that a = b. For example: sin( ) = sin( ) but .

To solve this problem of non 1-1'ness we simply erase almost all of the graph! I.E. we restrict the domain of f(x) = sin(x) to the interval on which the function is 1-1 and so has an inverse. We simply make sure to define the inverse g in such a way that it satisfies f(g(x)) = g(f(x)) = x. We call g(x), or "sine inverse of x". See blackboard for notation.

Definition : <=> for y and x .

*** Check that the above definition satisfies the composition requirements for inverse functions.

> sin(arcsin(x));arcsin(sin(x));

> [arcsin(-1),arcsin(0),arcsin(1/2),arcsin(1/sqrt(2)),arcsin(1)];

Simplify i) cos(arcsin(2/7)) ii) cos(arcsin(x))

> cos(arcsin(2/7));cos(arcsin(x));

> plot(arcsin(x),x=-Pi/2..Pi/2,colour=black);

> plot({sin(x),arcsin(x),x},x=-Pi/2..Pi/2,colour=[black,red,magenta]);

e.g. Find the derivative of y = arcsin(x) and generalize.

> d:=diff(arcsin(x),x);

Theorem : . Chain rule generalization: OBVIOUS!

> eg1:=arcsin(x^2)+(arcsin(x))^2;

> diff(eg1,x);

The Others.

Arccos(x)

> plot(cos(x),x=-20..20,colour=black);

To solve the non 1-1'ness of y = cos(x) we restrict the domain to and say:

definition : <=> for y and x [-1,1] .

You should check that the definition satisfies the composition equations.

> arccos(-1);arccos(0);arccos(1/2);arccos(1);

Simplify arccos(tan(3/a) if a > 0.

> assume(a>0);tan(arccos(3/a));

> simplify(%);

> plot({arccos(x),cos(x),x},x=-1..Pi,colour=[black,red,magenta],numpoints=1500);

Arctan(x)

> plot(tan(x),x=-10..10,y=-10..10,discont=true,colour=black);

To solve the non 1-1'ness of y = tan(x), restrict the domain to and say:

definition : artan(x) = y <=> x = tan(y) for y and x R .

Simplify i) sec(arctan(5/3)) ii) sec(arctan(x/3)).

> sec(arctan(5/3));sec(arctan(x/3));

> plot({arctan(x),Pi/2,-Pi/2},x=-10..10,y=-2..2,colour=black);

> Limit(arctan(x),x=infinity)=limit(arctan(x),x=infinity);Limit(arctan(x),x=-infinity)=limit(arctan(x),x=-infinity);

Now, use implicit differentiation to find the derivatives of the arcosine and arctangent functions.

> Diff(arccos(x),x)=diff(arccos(x),x);Diff(arctan(x),x)=diff(arctan(x),x);

Differentiate the following expressions:

> e1:=arctan(x^3);e2:=arccos(sqrt(2*t-1));e3:=(arctan(x))^(-1);

>

> diff(e1,x);diff(e2,t);diff(e3,x);

#65, pg 234. A ladder 10 ft long leans against a vertical wall. If the bottom of the ladder slides away from the base of the wall at a speed of 2 ft/s, how fast is the angle between the ladder and the wall changing when the bottom of the ladder is 6 ft from the base of the wall?

> equ:=cos(alpha)=x/10;

> equ2:=solve(equ,alpha);

> diff(equ2,x);

Now, if we consider x and to be functions of t where t stands for time then we have = . The problem gives us = 2 for all x, therefore, = .

> alpha_rate:=-2/sqrt(100-x^2);

To get the rate of change of w.r.t. x when x = 6 ft we plug in:

> subs(x=6,alpha_rate);simplify(%);

=> The angle is changing at the (instantaneous) rate of radians per second when the foot is 6 ft from the wall. Interprete.

> implicitplot(equ,x=0..10,alpha=0..2,colour=black);