Holiday Quiz!
Must Show all Work!
Neatness Counts!
1. Find the equation of the tangent
Line to the graph of y = secxtanx
at the point with x = .
Write the answer in exact form!
The point on the graph is (, sectan) = (, ) and the slope is y’(),
where y’ = [secxtanx](tanx) + (secx)[sec2x] = secxtan2x + sec3x, so
m = sectan2+sec3= =
So, with point (, ) and slope m = we can find the equation of the tangent line.
y - = (x-) or y = x- or y = x-
2. Sketch the graph including all intercepts, asymptotes and extrema:
y = =
We have intercepts:
(-2,0), (2,0) and (0, 2) and asymptotes y = ½, x = -1, and x = 1.
3. Same for: y = for x in [.
y’ =
0 =
which has no real solns
there are no critical numbers
Also note that y’ is always positive so that the graph must be increasing throughout the interval [. The graph also has asymptotes at the same values as tanx which are /2 and 3/2.
4. Same for: y = 4x3 – x2.
To find intercepts: 0 = x2( 4x – 1) (0,0) and (1/4, 0)
To find extrema: y’ = 12x2 – 2x = 2x(6x – 1) x = 0 and x = 1/6 are the critical numbers.
By the first derivative test, we have a local maximum at x = 0 and the maximum is 0.
Also, we have a local minimum at x = 1/6 and the local minimum is -0.009
Note that the values are very small and we need to zoom in on the graph to see what is happening.
5. Find the limits using limit properties:
a) b) c)
Since this is a limit at ∞,
We divide both the num. and = =
den by the greatest power of x
in the den. which is just x:= -∞ * (4-15) = ∞
=
=
= - ∞
6. Two commercial airplanes are flying at 40,000 ft along straight-line courses that intersect at right angles. Plane A is approaching the intersection point at a speed of 442 knots (nautical miles per hour; a nautical mile is 2000 yd). Plane B is approaching the intersection at 481 knots. At what rate is the distance between the planes changing when plane A is 5 nautical miles from the intersection point and plane B is 12 nautical miles from the intersection point? (Related Rates)
Given:
Find: when the distance from P to boat A is 5 knots and the distance from P to boat B is 12.
General equation: x2 + y2 = z2
7. A rectangle is to be inscribed under the arch of the curve y = 4cos(0.5x) from x = -to x = . What are the dimensions of the rectangle with the largest area, and what is the largest area? (Optimization)
The function to be maximized is the area, A.
A = 2x(4cos0.5x) = 8xcos(0.5x)
A’ = 8cos(0.5x) - 4xsin(0.5x)
To find the critical points we must use Newton’s method on A’, which yields the approximation formula:
xn+1 = xn - = xn -
To find the first estimate(s) of these critical points we must check the changes in sign of
A’:
A’(0) = 8
A’() = -12 **This means that there is a critical point between 0 and **
A’(2) = -8
A’(3) = + **This means that there is a critical point between 2 and 3**
A’(4) = +
A’(5) = - **This means that there is a critical point between 4 and 5**
A’(6) = -
A’(7) = + **This means that there is a critical point between 6 and 7**
A’(8) = +
This means that we must use Newton’s Method 4 times to get the 4 critical points.
CPT#1
For the first critical point I use 0.5 as my first estimate and then apply the formula (*) to get the first critical point to be 1.7206.
CPT#2
For the 2nd critical point I use 2.5 as my first estimate and then apply the formula (*) to get the 2nd critical point to be 6.85.
CPT#3
For the 3rdcritical point I use4.5 as my first estimate and then apply the formula (*) to get the 3rd critical point to be 12.875.
CPT#4
For the 4th critical point I use 6.5 as my first estimate and then apply the formula (*) to get the 4th critical point to be 19.058.
First Derivative Test:
We can see that the local maximums are (1.7206, 8.978), and (192.875, 101.78)
Since this problem is on a closed interval we must also check the endpoint x = 8:
A(8) = 201.06.
After all of this work, the maximum area occurs at x = 8!
8.What are the dimensions of the lightest open-top right circular cylinder can that can hold a volume of 1000 cm3?
V = 1000 cm3 = r2h h =
Function to be minimized is the Surface Area, S = r2 + 2rh which must be rewritten as a function of 1 variable: S = r2 + 2r = r2 + 2000r-1
Now we can differentiate S with respect to r:
Dr = 2r – 2000r-2 = 2r-2(r3 – 1000) We have critical numbers 0 and
By the First Derivative Test we have a minimum at r = and h =
9. Evaluate: , let u = , then du =
=
10.Evaluate: , let u = sinx, then du = cosxdx
=
11. Evaluate: , let u = tanx, then du = sec2xdx
=
12. Evaluate: =
13. Find the critical numbers of f(x) =
f’(x) = [18x2]+ (6x3)[] = 18x2-6x4
*which we can now factor*
=6x2(3( = 6x2(-4x2+24) = -24x2(x2 – 6)
=
So that we get critical numbers x = 0, x = , and x = with the last 2 being of type II.
14. Find the critical numbers of f(x) =
f’(x) =
=
From the derivative it appears that we have cn’s 0 and .
Upon inspection we see that are not in the domain so that the only cn is x = 0 which is
of type II.
15. Evaluate: = = tanx + secx + C
16. Find the volume of the solid of revolution obtained by revolving the region enclosed by the graphs of y = 2x2 – 2 and y = 3x+3 about the line x = 3.
First let us find the points of intersection: 2x2 – 2 = 3x + 3 2x2 – 3x – 5 = 0
(2x – 5)(x + 1) = 0 x = 5/2 and x = -1.
It is easiest to use cylindrical shells:
The radius of each cylindrical shell is r = 3 – x the height is the distance between the two
graphs, h = (3x+3) – (2x2 – 2) = -2x2 + 3x + 5.
V = 2 = . . . 202.04 cubic units
17. Same region as #16 revolved about the line y = -4.
Here it is best to use washers:
Ro is the distance between thethe axis of revolution and farthest function bounding the region
(the line):
Ro = (3x + 3) – (- 4) = 3x + 7
RI is the distance between the axis of revolution and the closest function bounding the region
(the parabola):
RI = (2x2 – 2) – (-4) = 2x2 + 2
V = = . . . 610.62 cubic units