Download Answers Test Calculus for AE (wi1276LR)

Delft University of Technology
Faculty of Electrical Engineering, Mathematics and Computer Science
Mekelweg 4, Delft
Answers Test Calculus for AE (wi1276LR)
Wednesday, October 29, 2008, 9:00 - 10:30 a.m.
1. H. Suppose that x = arcsin( 45 ), then we have: sin x = 54 and x ∈ [− π2 , π2 ]. In fact it
is easily seen that 0 < x < π/2. Now we p
look for: sin(2x) = 2 sin x cos x. We have
that: sin2 x + cos2 x = 1. Hence: cos x = ± 1 − (4/5)2 = ±3/5. Since x ∈ [− π2 , π2 ] we
conclude that cos x = 3/5 ≥ 0. Hence: sin(2x) = 2 sin x cos x = 2 · 45 · 35 = 24
25 .
2. A. An equation of the tangent line at the point (−1, 1) has the form y − 1 = r(x + 1),
dy . Using implicit
where r denotes the rate of increase. Now we have r = dx
(−1,1)
differentiation we obtain:
3x2 + 6xy + 3x2
dy
dy
+ 6y 2
=0
dx
dx
dy
3x2 + 6xy
x2 + 2xy
=− 2
=
−
.
dx
3x + 6y 2
x2 + 2y 2
=⇒
This implies that r = 1/3.
3. G. Using integration by parts we find that
Z
arctan x dx = x arctan x −
= x arctan x −
Z
x d arctan x = x arctan x −
Z
x
dx
1 + x2
1
ln(1 + x2 ) + C.
2
4. F. Using the substitution ln x = t we obtain
e4
Z
e
dx
√
=
x ln x
Z
e4
e
d ln x
√
=
ln x
Z
4
1
√ 4
dt
√ = 2 t = 2(2 − 1) = 2.
1
t
5. H. Note that the integrand becomes infinite for x = 0 ∈ (−1, 2). Further we have:
Z
0
2
dx
= lim
↓0
x4
Z
2
dx
1
= lim − 3
↓0
x4
3x
Hence the integral is divergent.
2
= lim −
↓0
1
1
= ∞.
+
24 33
6. F. The differential equation is separable:
(1 + y 2 ) dy = cos x dx
=⇒
1
y + y 3 = sin x + C.
3
From y(0) = 1 it follows that: C = 4/3.
7. A. The differential equation is linear. Look for an integrating factor I(x):
I(x)y 0 (x) + 2I(x)y(x) = 6I(x)ex .
For I(x) we must have that: I 0 (x) = 2I(x) which implies that I(x) = e2x (for instance).
Then we have:
i
d h 2x
e y(x) = 6e3x
dx
e2x y(x) = 2e3x + C
=⇒
8. F. In the polar form we have:
√
=⇒
y(x) = 2ex + Ce−2x .
√
3 − i = 2e−iπ/6 en 1 − i 3 = 2e−iπ/3 . Then we have:
√
10
( 3 − i)10
210 e− 6 π i
πi
5 ( 53 − 10
6 )
√
=
=
2
e
= 25 e0 = 32.
5
−3π i
5
(1 − i 3)5
2 e
9. B. The auxiliary equation is: r2 − 4r + 5 = 0 or equivalently (r − 2)2 + 1 = 0. This
implies that: r = 2±i. Hence: yh (x) = c1 e2x cos x+c2 e2x sin x. For a particular solution
we then have: yp (x) = Ae−x . Substitution leads to: 10A = 5 or equivalently A = 1/2.
Hence the general solution is: y(x) = yp (x) + yh (x) = 21 e−x + c1 e2x cos x + c2 e2x sin x.