Math 1A – Chapter 4 Test II Solutions – Fall ’04.

 

1.      Blanca is running around a circular track of radius r = 30 meters at a speed of 3 meters/sec.  Recall that the linear speed v is related to the angular speed ω by v = ωr.
 

a.       What is Blanca’s angular velocity, ω?
SOLN:  ω = 1/10 radians per sec.

b.      Recall that Blanca’s position at time t can be modeled by the  parametric equations .  If Blanca starts at a bench on the east side of the track and heads north around the track, what is a parameterization of her position at time t?
SOLN:

c.       Write a formula for the square of Blanca’s distance from the bench at time t.
SOLN: 

d.      Find the rate of change of Blanca’s distance from the bench when t = 1.
SOLN:  Differentiating the formula for D2: .  When t=1,  m/s.

 

 

2.      A paper cup has the shape of a cone with height 8 cm and radius 2 cm (at the top).  If water is poured into the cup at a rate of 2 cm3/s, how fast is the water level rising when the water is 6 cm deep?

SOLN:  .  By similar triangles, h = 4r.  Thus  and so .  Now when h = 6, r = 3/2. and thus  cm/sec.

 

3.      Find the point on the ellipse  which is closest to the point (3,5).
SOLN:  An arbitrary point in the first quadrant of the ellipse can be written as .  The square of the distance from this point to (3,5) is
Differentiating wrt x, .
Thus .  Equating squares, clearing fractions and expanding,
 



 Using the Voyage 200 to plot this function on the interval [0,1] and to find its root, we have (as shown in the graphic) x = 0.6064733.  This corresponds to the point (0.606,0.562) on the ellipse.

4.      An observer stands at a point P, 10 meters above a track.  Two runners start at a point S in the figure and run along the track.  One runner runs twice as fast as the other.  Find the maximum value of the observer’s angle of sight θ between the runners.
SOLN:  It helps to “think inside the box” for this one.  Note that if the slow runner has run a distance x, then the faster runner has run 2x and

that the sum of the angles is .  It’s easy to see that  and  so that .  Differentiating and setting , .  Thus .

You can avoid using β by looking at

Differentiating,
This problem can also be solved using the law of cosines:

 whence .  Differentiating, , we have
Plugging back into

 

5.      Use Newton’s method to find all solutions to the equation  correct to twelve decimal places.
SOLN:  The function  is zero when the equation is true.  Newton’s method iterates . 
On a programmable calculator, you can write a program to do Newton’s method. This is how that works on the TI Voyage 200:


This indicates that there are two solutions.  If the initial value is –1, the method converges to
–1.139615365664, while the initial value 1 leads to 0.650561444008

6.      Consider  

a.       Explain why g satisfies the conditions of the mean value theorem on .
SOLN: Since g is a product of a differentiable function with a composition of differentiable functions, it is differentiable everywhere—in particular, it is differentiable on .

b.      Set up an equation to find the point(s) in  which, by the mean value theorem are guaranteed to exist.
SOLN:

c.       Use Newton’s method to find a solution to the equation you set up.
SOLN:  A solution to the equation is a zero of .
Again, the Voyage 200 will get the job done. Note that the program has been modified to exit if the difference in iterates is just small rather than 0:

So with the initial value x1 = 0.5 the iterates converge to 1.77245385091 (which so happens to be the right endpoint of  …not on the interior. With the initial value x1 = 0.8 however, we get one of the two values whose existence is guaranteed by the theorem: approximately 0.9798922767

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