CHAPTER 2 - Derivatives

Section 2.1, page 117

Problem 51

Since the derivative of a quadratic function is linear, we know that there is a point on the parabola with slope -1, so we set about finding it:

>	y:=2*x^2-13*x+5;

>	solve(diff(y,x)=-1,x);

So the parabola has slope -1 when x=3. Now we can find the equation of the tangent line and graph - note the use of the "tangent line idiom":

>	tanline:=subs(x=3,y)+subs(x=3,diff(y,x))*(x-3);

>	plot({y,tanline},x=-1..6,color=blue,thickness=2);

Problem 60

We'll graph  thicker than the others so we can get an idea of what's happening:

>

f:=3*x^2;

>	f1:=((x+h)^3-x^3)/h;

>	with(plots,display):

>	A:=plot(f,x=-2..2,-0.1..3,thickness=4,color=blue):

>	B:=plot({subs(h=2,f1),subs(h=1,f1),subs(h=0.2,f1)},x=-2..2, -0.1..3,thickness=2,color=red):

>	display({A,B});

As h gets smaller the graphs of f1 approach that of f - as to be expected since the difference quotient in f1 is approaching the derivative of , which is f= .

From the other side:

>	C:=plot({subs(h=-2,f1),subs(h=-1,f1),subs(h=-0.2,f1)},x=-2..2,-0.1..3,thickness=2,color=red):

>	display({A,C});

Section 2.2, page 129

Problem 24

This is no problem:

>	diff((5*x+1)/(2*sqrt(x)),x);

>	simplify(%);

Problem 43

We'll use the "tangent line idiom" from before:

>	y:=4*x/(x^2+1);

>	tanline0:=subs(x=0,y)+subs(x=0,diff(y,x))*(x-0);

>	tanline1:=subs(x=1,y)+subs(x=1,diff(y,x))*(x-1);

Note that this second tangent line is horizontal. Now we can plot the curve and the two tangent lines:

>	plot({y,tanline0,tanline1},x=-4..4,-4..4,thickness=2,color=blue);

Section 2.3, page 139

Problem 13

We define the functions for the bullets on the moon and the earth:

>	moon:=832*t-2.6*t^2; earth:=832*t-16*t^2;

To find how long the bullets are in the air, solve for when their heights (these functions) are zero (of course, t=0 is an extraneous solution):

>	solve(moon=0,t); solve(earth=0,t);

So the lunar bullet is in the air for 320 seconds, and the terran one for 52 seconds.

To find the maximum height, set the derivative (velocity) equal to zero for the time, then plug into the height expression:

>	subs(t=solve(diff(moon,t)=0,t),moon);

>	subs(t=solve(diff(earth,t)=0,t),earth);

So the lunar bullet goes 66,560 feet up from the surface, and the terran bullet goes up only 10,816 feet.

Section 2.4, page 152

Problem 43

A quick limit problem for Maple:

>	Limit(sin(1-cos(t))/(1-cos(t)),t=0)=limit(sin(1-cos(t))/(1-cos(t)),t=0);

Problem 54

>	y:=2*x+sin(x);

The graph will have a horizontal tangent exactly where the derivative of this function is equal to zero.  Thus, we need to solve for where the derivative is equal to zero.

>	solve(diff(y,x)=0,x);

We have a little problem here, since arccos(2) does not exist (as a real number, at least). Just for the record, the derivative of y is

>	diff(y,x);

so we can see why this expression will never be equal to zero.  In particular, this tells us that the tangent line will never be horizontal.  We can now graph the function and see that this is indeed the case.

>	plot(y,x=0..2*Pi,color=blue,thickness=2);

Section 2.5, page 160

Problems 27 and 42

These are easy with Maple:

>	diff(1/21*(3*x-2)^7+1/(4-1/(2*x^2)),x);

>	diff((1+cot(t/2))^(-2),t);

>	simplify(%);

Section 2.6, page 170

Problem 42

To do implicit derivatives, you have to be careful to let Maple know that you are thinking of y as a function of x (by writing y(x)):

>

restart;

>	eqn:=x*y(x)+y(x)^2=1;

>	diff(eqn,x);

So Maple can take the derivative of both sides of an equation. To get the first and second derivatives of y, we have to solve this equation (and its derivative) for y' and y'':

>	yprime:=solve(diff(eqn,x),diff(y(x),x));

Now we can take the derivative of y' to get y'':

>	ypp:=simplify(subs(diff(y(x),x)=yprime,diff(yprime,x)));

Section 2.7, page 176

Problem 32

Let us call the distance from the car to the point on the track directly in front of you x.  Thus, we know that  feet per second, and . We're thinking of both x and  as functions of t, so we can write:

>	eqn:=tan(theta(t))=x(t)/132;

We want to compute d/dt at the moment when x=0, and a half second later (when we will

have x=132 feet).

We differentiate both sides of the second equation above and get

>	diff(eqn,t);

We need to solve this for d/dt --

>	thetaprime:=solve(diff(eqn,t),diff(theta(t),t));

But we know that , so we can substitute that into what we just calculated:

>	thetaprime:=subs(tan(theta(t))=x(t)/132,thetaprime);

Now we can calculate the derivatives we need:

>	subs(diff(x(t),t)=264,x(t)=0,thetaprime);

>	subs(diff(x(t),t)=264,x(t)=132,thetaprime);

So the camera angle is changing at a rate of 2 radians/sec when the car is in front of us, and at a rate of 1 radian per second a half-second later.