Calculus: The Language of Change

Exercise Set 8.2
e as a "Natural" Base

by Sharon Portz

Exercise 1)

part a

Let y = (b^t) for an unknown (but fixed) positive constant b.  Use the Chain Rule (see Section 6.5) to show that y[t] satisfies

y[0] = 1

(dy/dt) = k y

What is the value of the constant k?

Solution  to first part of a

To answer the first part we can see:

y[t] = (b^t)  and

y[0] = (b^0) = 1

Now, to answer the second part we look to the text.

The text on page 99 reads:

If we want to differentiate y = (b^t) we can use the preceding Chain Rule computation and two other important facts about exponentials.  We know the general exponential law

((((a^c)))^x) = (a^cx)

so we try to find a constant c that satisfies

2 = (e^c)

Once we find this c we have (2^t) = (e^ct) for all t because of the law of exponents.  We know how to differentiate y = (e^ct), and with this value of c, we learn how to differentiate (2^t),

((d((2^t)))/dt) = ((d((e^ct)))/dt) = c (e^ct) = c (2^t).

Now we solve for c using natural log.  Nataural log and exponential are inverse functions.  This simply means

Log[(e^t)] = t 	and 	\(TraditionalForm\`\(	e\^Log[s]\)\) = s,  s>0

We apply this to our problem by taking logs of both sides of

2 = (e^c)

Log[2] = Log[(e^c)] = c

Thus c = Log[2] and we see that

((d((2^t)))/dt) = Log[2]  x (2^t)

In general,

((d((a^x)))/dx) = Log[a] (a^x)

Solution to the second part of a

Using this method, we can find ((d((b^t)))/dt).  It is:

(dy/dt) = ((d((b^t)))/dt) = Log[b](b^t)

Since  y=(b^t) we have

(dy/dt) = ((d((b^t)))/dt) = Log[b]y

Let k = Log[b] and we have

(dy/dt) = ((d((b^t)))/dt) = k y

Therefore  the value of the constant k is Log[b].

part b

Show that  y = (e^kt) satisfies

y[0] = 1

(dy/dt) = k y

If the constant  k is the same as in the first part, how much is (e^k) in terms of b?

Solution of part b

y[t] = (e^kt)

y[0] = (e^k0) = (e^0) = 1

(dy/dt) = ((d((e^kt)))/dt) =\(d[e\^k]\^t\/dt\)

		= Log[(e^k)] (e^kt)
		
		= k  (e^kt) = k y

Now, if k = Log[b] we have

		(dy/dt) = Log[b] (e^kt)
		
so		Log[(e^k)] = Log[b]

and		b = (e^k)

part c

Solve the initial value problem

y [0] = 5

(dy/dt) = y

Solution of part c

For y = k (e^t) we know that (dy/dt) = k (e^t) = y


Since (e^0)= 1, in order for y[0] = 5 we must have k = 5 so our solution is

		y[t] = 5 (e^t)

part d

Solve the initial value problem

y[0] = 5

(dy/dt) = k y

where k = Log[2].  Show that your solution may also be written as y = 5*(2^t)

Solution of part d

For y = c (e^(k t)) 

		y[0] = c
		
and		(dy/dt) = k c (e^(k t))

So we need	c = 5

which gives us
		y =5 (e^(k t))
		
and		(dy/dt) = 5 k (e^(k t)) = k y

We also want k = Log[2] so

		y = 5 (e^(Log[2] t))
		
Since (e^Log)[2] = 2, this is equivalent to

		y =5*(2^t)

moral of the exercise

The moral of this exercise is this: We could write solutions of initial value problems

y[0] = (Subscript[Y, 0])

(dy/dt) = k y

as y = (Subscript[Y, 0])* (b^t), where b = (e^k) for Log[b] = k; but, for the purposes of calculus, it is "more natural" to write them  in the form y = (Subscript[Y, 0])*(e^kt),

y[0] = (Subscript[Y, 0])

dy = k y dt

->	y[t] = (Subscript[Y, 0])*(e^kt),  t ϵ [0,∞)