MAT1320B Final Examination 2018
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MAT1320B Final Examination 2018
[3] 1. Let f (x) = (x + 1)2 . Find the derivative f′ (x) directly from the definition. You must find it
using the definition, not some other method.
solution:
d (x + h + 1)2 ± (x + 1)2
dx h→0 h
= lim
= lim
h→0 1
= 2x + 2
2. Give the derivative f′ (x) of each function. For this question, no justification required.
a) f (x) = arcsin(x)
d 1
dx ,1 ± x2
x
b) f (x) =
d (1)(3x + 1) ± (x)(3)
dx (3x + 1)2
c) f (x) = ex2
solution: f (x) = ex2 = ex2 2x
3. Using logarithmic differentiation, determine the derivative of g(x) = (x2 + 1)cos(x). Your final answer should be completely in terms of x.
solution:
ln(g) = ln ╱(x2 + 1)cos(x)、= cos(x) ln(x2 + 1)
= ± sin(x) ln(x2 + 1) + cos(x) (2x)
g ′ = g(x) ;± sin(x) ln(x2 + 1) + cos(x) (2x)、 = (x2 + 1)cos(x) ;± sin(x) ln(x2 + 1) + cos(x) (2x)、
[4] 4. Consider the curve described by x5 + xy4 + y2 = 1.
dy
solution:
╱x5 + xy4 + y2、= (1)
5x4 + 1y4 + x4y y + 2yy3′ ′ = 0
╱ 5x4 + y4、+╱4xy3 + 2y、y′ = 0
dx = y = 4xy3 + 2y
b) Find every point on the curve that has x = 0, and the slope of the curve at each such point.
solution: Setting x = 0 in the equation of the curve we get (0)5 + (0)y4 + y2 = 1, which gives y2 = 1 so y = ·1. We have two points: (0, 1) and (0, ± 1).
±5(0)4 ± (1)4 1
4(0)(1)3 + 2(1) 2
±5(0)4 ± ( ± 1)4 1
4(0)( ± 1)3 + 2( ± 1) 2
[3] 5. A ladder of length 10 metres leans against the wall, making an angle θ with the ground. The
ladder slides down the wall, and when θ = π/3 then θ is decreasing at a rate of 0.01 radians/sec. At this moment, how fast is the base of the ladder moving away from the wall?
solution: (picture omitted:)
Let x be the horizontal distance from wall to ladder, and θ the angle between the ground and
dx dt dθ
[3] 6. For each f (x), give an anti-derivative F (x). Find an anti-derivative that is valid on the whole
domain of the function. You do not need to include the constant of integration, +C. For this question, no justification required.
a) f (x) = x5 +,x
solution: +
b) f (x) = sec2 (x)
solution: tan(x)
c) f (x) =
solution: 2 ln lx + 1l
[2] 7. You are given that g′ (t) = t2 + sin(t) and that g(0) = 0. Find the function g(t).
solution: We want an antiderivative of g′ . So we get g(t) = t3 /3 ± cos(t)+C for some constant C. setting 0 = g(0) = 03 /3 ± cos(0)+C = ± 1+C we see that C = 1. So g(t) = t3 /3 ± cos(t)+1.
d x2 cos(t)
dx 6 ln(t + 2)
solution: dt = (2x).
[6] 9. Calculate each definite integral. Show your steps!
3
a) et cos(et ) dt 2
solution: Substitute w = et , dw = et dt.
2 3 et cos(et ) dt = cos(w) dw = ╱sin(w)← │e(e)2(3) = sin(e3 ) ± sin(e2 )
b)
e′x dx
1
solution: Substitute z = ,x, and dz = dx/(2,x). This means that dx = 2,x dz = 2z dz
4 2
Now we can do this by parts with u = z, dv = ez dz. This gives du = dz and v = ez .
2 1 2 ez z dz = 2 ╱ ╱zez ← │ ez dz\ = 2 ╱ ╱zez ← │1(2) ± ╱ez ← │1(2)\
= 2 ╱(2e2 ± e) ± (e2 ± e)、 = 2e2
[4] |
10. Using the method of partial fractions (and not some other method), calculate Be sure and give your final answer in terms of the original variable. |
|
dx. |
solution:
1 a bx + c
x(x2 + 1) x x2 + 1
1 = a(x2 + 1) + (bx + c)x
Setting x = 0 gives a = 1. Setting x = 1 gives 1 = 2a + b +c = 2 + b +c. Setting x = ± 1 gives 1 = 2a + b ± c = 2 + b ± c. Solving this gives c = 0 and b = ± 1. Now we can integrate.
1 dx =
1 dx +
±x
dx
The second integral can be done with u = x2 + 1 and du = 2x dx
dx = ln lxl + ±
du
= ln lxl + ± ln lul
= ln lxl + ± ln │x2 + 1 │ + C
11. Using the method of trig substitution (and not some other method), calculate Be sure and give your final answer in terms of the original variable.
,1 ± x2 dx.
solution: We set x = sin θ, and so dx = cos θ dθ .
,1 ± x2 dx = 尸1 ± sin2 (x) cos θ dθ = cos2 θ dθ
Apply an identity, integrate, and apply another identity.
cos2 θ dθ = ╱ 1 + cos(2θ)← dθ = ╱θ + sin(2θ)← + C = ╱θ + 2 sin θ cos θ←
In order to transform back we’d draw a triangle that tells us that when sin θ = x, then cos θ = ,1 ± x2 . So we get the final answer.
;arcsin(x) + 2x,1 ± x2 、+ C
1
[4] 12. We wish to evaluate ex2 dx numerically. In case it is useful, you are given the following
0
derivatives for f (x) = ex2 .
f (x) = 2xe′ x2 f′′′ (x) = ╱8x3 + 12x、ex2
f′′ (x) = ╱4x2 + 2、ex2 f′′′′ (x) = ╱ 16x4 + 48x2 + 12、ex2
a) Give the expression for the Riemann sum using the right hand rule with n = 4. You do not need to evaluate it, just give the sum.
solution: ;e(0.5)2 + e(1)2 + e(1.5)2 + e(2)2 、(0.5)
b) Give the expression for the approximation using Simpson’s method with n = 4. You do not need to evaluate it, just give the sum.
solution: ;e(0)2 + 4e(0.5)2 + 2e(1)2 + 4e(1.5)2 + e(2)2 、(0.5)
c) Determine the value of n required so that the trapezoid method is accurate to within 0.001.
solution: We want lf′′ (x)l < K. Note that f′′ (x) is positive. Also it is increasing (since f′′′ (x) > 0). So the maximum of lf′′ (x)l occurs at the right-hand endpoint, namely x = 1. So we set K = lf′′ (1)l = 6e.
Now we want < 0.001. This gives n 2 ! .
[4] 13. Find the absolute maximum and absolute minimum value and the location(s) where they
ln(x)
2
dg (1/x)x2 ± ln(x)2x 1 ± 2 ln(x)
dx x4 x3 .
We find that f′ (x) is positive on [1/e, e) and negative on (e, e2]. This means there is an absolute maximum at x = e, giving f(e) = ln(x)/x2 = 1/e2 .
The minimum is at one of the endpoints (since there are no local minima inside the interval.
f(1/e) = = = ±e2
f(e2 ) = = =
The former is negative and the latter positive, so the minimum is ±e2 , which occures at x = 1/e.
[4] 14. Find each of the following limits. You may use any method we have covered in this course.
et + 1
t→& t2 + t
solution:
lim et + 1 lim et lim et
This limit now tends to o.
b) lim(x + 1)1/x x→0
solution:
exp ;ln ╱x(l)(x + 1)1/x←、= exp ╱x(l) ln ╱(x + 1)1/x、←
= exp ╱x(l) ln (x + 1)\
= exp ╱x(l) \
╱ \
= exp ╱1、
= e
[5] 15. Consider the following function and its derivatives.
x3 x2 (x2 ± 3) 2x(x2 + 3)
x2 ± 1 (x2 ± 1)2 (x2 ± 1)3
a) Identify all horizontal and vertical asymptotes.
solution: Check limits:
lim = o
lim = o
lim = o
solution: Critical points are at 0, ·1, ·,3.
±,3 ± 1 0 1 ,3
+ 0 ± ! ± 0 ± ! ± 0 +
x y y y y x
There is a local maximum at x = ±,3 and a local minimum at x = ,3.
c) Determine where the function is concave up and where the function is concave down. Identify all inflection points.
solution: Critical points are at 0, ·1.
± 1 0 1
±
n
!
+ 0 ± ! +
u n u
There is an inflection point at x = 0.
d) Sketch the curve, indicating the special points found above.
solution:
[+3] 16. (bonus) You are given a square piece of cardboard, 1 metre on a side. You are to cut out a
square of side length x from each corner (where x is to be determined), and then fold up the sides in order to make a box with an open top. Find the largest volume your box could have.
1m
cut out
corners
x x |
||
x |
x |
|
x |
x |
|
x x |
fold on
dotted lines
|
x x |
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x x |
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x x |
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x x |
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solution: Box has a base of 1 ± 2x by 1 ± 2x; the height is x. The domain of is x e (0, 1/2).
2022-04-14