CHAPTER 3 - Applications of Derivatives

Section 3.1, page 195

Problem 10

To find absolute extrema on closed intervals, check where the derivative is 0, where it fails to exist, and at the endpoints:

>

y:=4-x^2;

>	dy:=diff(y,x);

>	solve(dy=0,x);

The derivative exists for all x, and so we have to check the interior critical point x=0 as well as the endpoints x=-3 and x=1:

>	[-3,subs(x=-3,y)],[0,subs(x=0,y)],[1,subs(x=1,y)];

We see that the minimum is -5 (at x=-3) and the maximum is 4 (at x=0).

Section 3.2, page 203

Problem 10

We need to choose a, m and b so that the function

>	f:=piecewise(x=0,3,0<x and x<1,-x^2+3*x+a,1<=x and x<=2,m*x+b);

satisfies the hypotheses of the mean-value theorem on [0,2]. So it has to be continuous on the closed interval [0,2] (in particular at x=0 and x=1), and differentiable on the open interval (0,2) (in particular at x=1). So we solve three equations for three unknowns:

>	solve({3=limit(-x^2+3*x+a,x=0,right),limit(-x^2+3*x+a,x=1,left)=limit(m*x+b,x=1,right),limit(diff(-x^2+3*x+a,x),x=1,left)=limit(diff(m*x+b,x),x=1,right)},{a,m,b});

>	f:=subs(%,f);

This function should be continuous on [0,2] and differentiable on (0,2) -- let's see:

>	plot(f,x=0..2,color=blue,thickness=2);

Section 3.3, page 208

Problem 21

Start by defining the function and its derivative:

>	g:=x*sqrt(8-x^2);

>	dg:=diff(g,x);

Of course, the domain of this function is where , i.e., for x between  and .

Let's see where the derivative is positive and negative:

>	solve(dg>0,x);

>	solve(dg<0,x);

So the function is increasing when x is between -2 and 2, and decreasing to the left of -2 and to the right of 2. We calculate the critical values:

>	simplify(subs(x=-2,g)),simplify(subs(x=2,g));

So the min point of the graph is at [-2,-4] and the max point is at [2,4]. These are absolute, since g is zero at the endpoints of the domain.

Now let's plot:

>	plot(g,x=-2*sqrt(2)..2*sqrt(2),color=blue,thickness=2);

Section 3.4, page 217

Problem 21

We use calculus to make sure we have all the "interesting" points in our Maple plot:

>	y:=x^5-5*x^4;

>	solve(y=0,x);

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

>	solve(diff(y,x$2)=0,x);

Since y is a polynomial, we don't have to worry about any asymptotes. We need only

plot on an interval that contains all of x=0,3,4,5. To get an idea of the range on the y-axis, we'll substitute our "interesting" x-values into the function:

>	subs(x=0,y),subs(x=3,y),subs(x=4,y),subs(x=5,y);

So we have to include a pretty big y range:

>	plot(y,x=-4..7,-350..100,color=blue,thickness=2);

Section 3.5, page 230

Problem 101

First we'll do the three graphs:

>	y:=-(x^3-2)/(x^2+1);

>	plot(y,x=-9..9,color=blue,thickness=2);

>	plot(y,x=-90..90,color=blue,thickness=2);

You can see a very tiny kink in the graph at the origin.

>	plot(y,x=-900..900,color=blue,thickness=2);

Now it just looks like the line y=-x. This is because:

>	simplify(y+x);

>	limit(y+x,x=infinity),limit(y+x,x=-infinity);

So the line y=-x is an oblique asymptote for the graph, and once you subtract the "dominant" -x from y, you get a function that goes rapidly to 0 as x gets large.

Section 3.6, page 242

Problem 30

Let x be the length of the beam, let h be the height on the building's wall where the beam touches, and let s be the distance from the little wall where the other end of the beam touches the ground. Then we want to minimize x (or, what is the same, ), where we have the following relationships between x, s and h:

>	eqn1:=x=sqrt(h^2+(27+s)^2); eqn2:=h/8=(s+27)/s;

The second equation is the "constraint", and the first provides our "objective function". So we solve the second for h in terms of s, and then substitute into the first:

>	eqn1:=subs(h=solve(eqn2,h),eqn1);

Now we can take the derivative and set it equal to 0:

>	solve(diff(rhs(eqn1),s)=0,s);

We certainly don't want an imaginary root! So the minimum must occur when s=12 feet. We can substitute this into the first equation to see what x should be:

>	simplify(subs(s=12,eqn1));

So the shortest beam that reaches is  feet long.

Problem 56

From the information given in the problem we have the following formulas, where x is the number of people who are on the tour.

>	cost:=6000+32*x;

Given that at least 50 people go on the tour, we have the following formula for revenue.

>	revenue:=x*(200-2*(x-50));

>	profit:=revenue-cost;

To maximize profit, we find where its derivative (also known as the marginal profit)  is equal to zero, and then we must check that this is in fact a maximum.

>	mp:=diff(profit,x);

>	solve(mp=0);

The critical point of the profit function is 67 people. Since the second derivative of profit is -4 (we don't need Maple to tell us that!), this must be a maximum for the profit. And the profit is:

>	subs(x=67,profit);

dollars.

Section 3.7, page 257

Problem 42

We have to calculate the differential of the surface area:

>	S:=Pi*r*sqrt(r^2+h^2);

>	subs(r=r[0],simplify(diff(S,r)))*dr;

This is how much the surface area changes if r changes from  to .

Section 3.8, page 266

Problem 19

First, let's plot y=tan(x) and y=2x to see that they do cross:

>	plot({tan(x),2*x},x=0..Pi/2,0..5,color=blue,thickness=2);

OK, now we have to set up Newton's method - we might as well start with the initial guess x=1.

>	f:=x->tan(x)-2*x;

>

p:=1.0;

>	p:=p-f(p)/D(f)(p);

>	p:=p-f(p)/D(f)(p);

>	p:=p-f(p)/D(f)(p);

>	p:=p-f(p)/D(f)(p);

>	p:=p-f(p)/D(f)(p);

>	p:=p-f(p)/D(f)(p);

>	p:=p-f(p)/D(f)(p);

It looks like we have converged to the answer. Just to check, let's use Maple's fsolve:

>	fsolve(f(x)=0,x=0..2);