Solutions, Test 3

Notational changes (due to the limitations of HTML):

• "Pi" stands for the constant pi = 3.14159265.....
• "sqrt" stands for "square root." That is, sqrt(N) means N^1/2

1. Tangent to f(x)=x² - 6x + 8, at x=2.

The slope of the tangent line at any value of x is given by the derivative of f(x).  In this case, f'(x)=2x-6, so the slope when x=2 is f'(2)=-2..  Therefore, the tangent line is of the form y=-2x+b and passes through the point (2,0) (since f(2)=0).  Substituting x=2 and y=0 into the equation of the tangent line, we find that b=4.  Therefore, the equation of the tangent line is y=-2x+4.

The slope of the tangent line is f'(x), evaluated at x=2.  Since f'(x)=-1/x², the slope is -1/4.  So, the equation of the line is y=(-1/4)x+b, and the line passes through the point (2,1/2).  Solving for b gives us b=1, so the equation of the tangent line is y=(-1/4)x+1.

The slope of the tangent line is f'(x), evaluated at x=Pi/4.  Since f'(x)=2cosx (-sinx) - 2sinx (cosx) = -4sinx cosx, and sin(Pi/4) = cos(Pi/4) = sqrt(1/2), the slope is f'(Pi/4)=-4(sqrt(1/2))² = -4/2 = -2.  Therefore, the equation of the line is y=-2x+b, and the line passes through the point (Pi/4, 0) (since f(Pi/4)=1/2 - 1/2 = 0).  Solving for b gives us b=Pi/2, so the equation of the tangent line is y=-2x+Pi/2.

Since f'(x)=(1/2)(x2+1)^-1/2(2x)=x/sqrt(x²+1), the slope is f'(sqrt(3))=sqrt(3)/sqrt(4)=sqrt(3)/2.  So, the equation of the line is y=(sqrt(3)/2)x+b, and the line passes through the point (sqrt(3),2).  Solving for b gives us b=1/2, so the equation of the tangent line is y=(sqrt(3)/2)x+1/2.

Since this is not written as a function of x, use implicit differentiation to find dy/dx, and then evaluate at x=1.  Differentiating both sides with respect to x gives us 2x + 2y(dy/dx) = 2; solving for dy/dx gives us dy/dx=(1-x)/y.  Therefore, when x=1, dy/dx=0.   Thus, the tangent line(s) will be horizontal.

To find the point(s) on the graph at which x=1, substitute x=1 into the equation of the graph and solve for y.  Doing so gives us y²=1, so there are two solutions: y=1 and y=-1.  Since each line is horizontal, the equations of the tangent lines are y=1 and y=-1.

(Note:  Actually, you should have been able to draw this graph, to help you visualize the problem.  This is because, if we subtract 2x from both sides and then complete the square in x and y, the equation becomes (x-1)²+y²=1.  This, of course,
is the equation for the circle with center (1,0) and radius 1.)

4. Find dy/dx on the curve 3x²-2y²=1.

Since this equation is not solved for y, it is easiest to use implicit differentiation. (Solving for y first also will work, but it is much more difficult.) Differentiating both sides with respect to x, we get:

6x-4y(dy/dx)=0

dy/dx = 6x/4y = 3x/2y.

III. An object's position at time t is given by the function x(t)=t³-2t².  Determine when, if ever, the object:

1. ...changes direction.

"Direction" is determined by the sign of the velocity (positive velocity means right, negative velocity means left), so we need to determine that value(s) of t for which t changes sign.  So first we find the velocity of the object:
 

v(t)	=	x'(t)
	=	d dt	(t3/6 - 2t2)
	=	(3t2)/6 - 2(2t)
	=	t2/2 - 4t
	=	1/2(t2-8t)

Therefore, v(t)=0 when t=0 or t=8.  Since the object isn't moving for t<0, we only need to be concerned with t=8.  By testing intervals, we find that v(t)<0 when 0<t<8, and v(t)>0 when t>8.  Therefore, the object does change its direction at time t=8.

2. ...is moving to the left, but slowing down.

This refers to both velocity and acceleration.  "Moving to the left" means v(t) is negative, while "slowing down" means v(t) and a(t) are of opposite sign -- in this case, then, a(t) must be positive.  We already know (from our work in problem #1) that v(t) is negative iff 0<t<8.  Now, we must find a(t):
 
 

a(t)	=	v'(t)
	=	d dt	(t2/2 - 4t)
	=	(2t)/2 - 4(1)
	=	t - 4

Clearly, a(t)>0 iff t>4.  Therefore, the object is moving to the left, but slowing down, iff 0<t<8 and t>4.   Combining these two conditions gives us the solution: the object is moving left, but slowing down, iff 4<t<8.

This problem is similar to #2: "moving to the right" means v(t)>0, and "speeding up" means a(t) is of the same sign as v(t), so a(t)>0.  From #1, we know that v(t)>0 iff t>8; from #2, a(t)>0 iff t>4.  Therefore, we need t>8 and t>4; combining these two conditions gives us the solution set t>8.

This allows one to quickly and easily determine the sign of v(t) and of a(t) during any interval of time.  (For more detailed information, we could graph v(t) and a(t) in two dimensions; however, that is not necessary for these problems, since all that matters are the signs of these two functions.)

IV. Galileo's formula for free fall is y(t)=-(1/2)gt²+v₀t+y₀.   An object is dropped froma height of 400 ft.

(Recall that, when measuring in feet, g=32 ft/_sec2.)

1. Find expressions for v(t) and a(t) (in terms of t).

First of all, note that the object's initial velocity is v₀=0, and its initial height is y₀=400.  Therefore,  the equation for the motion of this object will be y(t)=-16t²+400.  The functions v(t) and a(t) are the first and second derivatives of y(t), respectively:

v(t)	=	y'(t)			a(t)	=	v'(t)
	=	d dt	(-16t2 + 400)			=	d dt	(-32t)
	=	-32t				=	-32

Solution: v(t)=-32t, a(t)=-32

2. After how many seconds will the object hit the ground?

"The object hits the ground" is another way of stating that y(t)=0.  Solving this equation gives us:

0	=	-16t2+400
16t2	=	400
t2	=	25
t	=	5

Solution: The object will hit the ground after 5 seconds.

The object hits the ground when t=5, so all we need to do is evaluate |v(5)|.  (Note that "speed" is always positive, i.e. the absolute value of the velocity.  Since we know from #1 that v(t)=-32t, v(5)=-160.  Therefore, the object's speed when it hits the ground will be 160 ft/sec.

Note!  The value 16(Pi) is not dr/dt, nor is it dV/dr, nor is it any of the other "creative" interpretations I observed on many students' exams.   "Cubic centimeters" implies volume, and "per second" implies time, so what you are being given here is dV/dt -- the rate of change of the VOLUME with respect to TIME.

You're being asked to find dr/dt - the rate of change of the radius with respect to time.  We know that V=(4/3)(Pi)r³; differentiating both sides with respect to time gives us:
 

dV dt	=	d dt	((4/3)(Pi)r3)
	=	(4/3)(Pi)(3r2)(dr/dt)
	=	4(Pi)r2(dr/dt)

 

Therefore,

dr dt

=

(dV/dt) 4(Pi)r2

=

16(Pi) 4(Pi)r2

=

4 r2

(Reminder: at the beginning of the problem, we were given: dV/dt=16(Pi).)

The above equation tells us dr/dt for any positive value of r; substituting r=5 gives us:

 

Solution:

dr dt

=

4 25

feet per second

(Note: one could also use the chain rule:
 

dr dt

=

dV dt

dr dV

=

dV/dt dV/dr

,

which also leads to the formula for dr/dt which we found above by using implicit differentiation.)

You are being asked to find dS/dt when r=5.  Proceeding as in problem #1:

 

S	=	4(Pi)r2
dS dt	=	d dt	(4(Pi)r2)
	=	4(Pi)(2r)(dr/dt)
	=	8(Pi)r(dr/dt)

From the previous problem, we know that dr/dt=4/25 when r=5.
 

Therefore,

dS dt

=

8(Pi)(5)

4 25

=

160(Pi) 25

=

32(Pi) 5

square feet per second.

(Note that, since we're discussing area, the units are square feet.)

VI.  Use differentials to approximate each of the following, without the use of a calculator.

(Note: since there is no "delta" on the keyboard, I'll be using "D" in place of "delta."  That is, Dx means "delta x," and Df means "Delta f.")

1. 50^1/2

Let f(x) = x^1/2, x₀=49 and Dx=1.  Note that choosing x₀=49 makes sense because it's close to 50 and we know the exact value of f(49).
Then, f(x₀)=f(49)=7, and

Df	=	Dx*f'(x0)
	=	1*(1/14)	(since f'(x)=(1/2)x-1/2)
	=	1/14, or approximately 0.0714

(Note: since all we're looking for here is an approximation, rather than an exact solution, it's acceptable to use a decimal approximation in place of a fraction.) Therefore, f(50) is approximately f(49)+Df, which is 7 1/14 , or approximately 7.0714

Let f(x) = x^2/3, x₀=27 and Dx=30-27=3.  We choose x₀=27 because it's close to 30 and we know that f(27)=9.
Then, f(x₀)=f(27)=9, and

Df	=	Dx*f'(x0)
	=	3*(2/3)*(1/3)	(since f'(x)=(2/3)x-1/3)
	=	2/3, or approximately 0.6667

Therefore, f(50) is approximately f(27)+Df, which is 9 2/3, or approx. 9.6667.

For this problem, the best of the "special angles" to use as a starting point is Pi/4, since it's closer to 4(Pi)/15 than the others.  However, starting from Pi/6 or Pi/3 was also acceptable.

Let x₀ = Pi/4; then, Dx=4(Pi)/15 - Pi/4 = Pi/60.  Then, sin(x₀)=(1/2)^1/2, and

Df	=	Dx*f'(x0)
	=	(Pi/60)*(1/2)1/2	(since f'(x)=cos x)
	=	(Pi)*21/2 120 , or approximately 0.03702

Therefore, sin(4(Pi)/15) is approximately sin(Pi/4)+Df, which is (1/2)1/2+(Pi/60)*(1/2)1/2, or approximately 0.7441.

Extra Credit.

1. Let y=sin^-1x.  Find dy/dx in terms of x.  (Recall that sin^-1x denotes the inverse of the sine function, for -1<x<1, -Pi/2<y<Pi/2.

The important observation to make here is that x=sin y.  (That's what an "inverse of a function" means.)  Differentiating both sides with respect to x, we get:
 

x	=	sin y
1	=	cos y (dy/dx)
dy/dx	=	1/(cos y)

This gives us dy/dx, in terms of y.  However, the problem is to write this in terms of x.  In the interest of determining who is reading this, if anyone, I'll leave this problem open.  Whoever is the first student in the class to show my how to write dy/dx in terms of x will get 5 extra credit points.

Hint:  the final answer will not contain any trigonometric functions (believe it or not)!  Instead, you should get a rational function of x.

First be clear on what you are given, and what you are to show.

Given: f(x)=f(-x) , for any x such that x and -x are both in the domain of f.

Show:  f'(-x) = -f'(x)  (this is what it means for f'(x) to be an odd function.)

Proof:  The most direct approach is probably to differentiate both sides of the above equation, f(x)=f(-x), with respect to x:
 

f(x)		=	f(-x)
d dx	f(x)	=	d dx	f(-x)
f'(x)		=	f'(-x)*(-1)		(due to the chain rule!)
f'(x)		=	-f'(-x),		which was to be shown!  (QED)

For a proof of the Product Rule, see p. 144 of the textbook.