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DEPARTMENT OF MATHEMATICS, IIT - GUWAHATI
Odd Semester of the Academic year 2019 - 2020 MA 101 Mathematics I Problem Sheet 1: Revision of vectors, equations of lines and planes, vector differentiation, limits and continuities of functions of several variables. Instructors: Dr. J. C. Kalita and Dr. S. Bandyopadhyay
Vb
Vr
θ
θ x
y
V
x
y
A man wants to paddle his boat across a river from point X to the point Y on the opposite shore directly across from X. If he can paddle the boat at the rate of 5 kilometers per hour and the current in the river is 3 kilometers per hour, in what direction θ should he steer his boat in order to go straight across the river? Also what is his resultant speed across the river?
Solution: Let V (^) b be the velocity of the boat and V (^) r be the velocity of the river and let V be the resultant velocity of the boat across the river. Clearly if V (^) r = 3i then V (^) b = 5 cos(θ)i + 5 sin(θ)j and V = αj for some α > 0, where α = |V | is the relative speed of the boat across the river. Also V (^) b + V (^) r = V ⇒ (5 cos(θ) + 3)i + 5 sin(θ)j = αj (1) Equating the components of the vectors on both sides of equation (1) we get 5 cos(θ) + 3 = 0 ⇒ cos(θ) = −^35 ⇒ θ = 2.214 radians, or θ = 126. 87 o.
Since α^2 + 3^2 = 5^2 , α = 4, hence the resultant speed is 4 kilometers per hour.
- Use a scalar projection to show that the distance from a point P 1 (x 1 , y 1 ) to the line ax + by + c = 0 is |ax 1 + by 1 + c| √ a^2 + b^2
Use this formula to find the distance from the point (− 2 , 3) to the line 3x− 4 y+5 = 0.
Solution: Let P 0 (x 0 , y 0 ) be the foot of the perpendicular vector from P 1 (x 1 , y 1 ) to the line ax + by + c = 0, then distance of P 1 (x 1 , y 1 ) from the line ax + by + c = 0 is given by
∣∣ ∣
P 1 P 0
∣∣ ∣.
P (^ x^ ,y)
x 0 y 0
x 1 y 1
ax+by+c=
P 0 ( , )
P 1 (^ , )
Let P (x, y) be a point on the line ax + by + c = 0, then P 0 (x 0 , y 0 ) and P (x, y) satisfy the equations ax 0 +by 0 +c = 0 and ax+by+c = 0, respectively ⇒ a(x − x 0 ) + b(y − y 0 ) = 0, ⇒ 〈a, b〉 · 〈x 0 − x, y 0 − y〉 = 0. Hence 〈a, b〉 ⊥ 〈x 0 − x, y 0 − y〉 ( or 〈a, b〉 is orthogonal to 〈x 0 − x, y 0 − y〉 ) ⇒ 〈a, b〉 ⊥
P P 0.
But
P 1 P 0 ⊥
P P 0 therefore any nonzero vector α along
P 1 P 0 must be parallel to 〈a, b〉, and α = λ 〈a, b〉 for some λ ∈ R, λ 6 = 0. If we denote
P 1 P by β, then ∣∣ ∣
P 1 P 0
∣∣ ∣ =
∣∣ ∣projα(β)
∣∣ ∣ =
∣∣ ∣∣ ∣
α.β |α|
∣∣ ∣∣ ∣
=
∣∣ ∣∣ ∣
λ(a(x 1 − x) + b(y 1 − y)) |λ|
a^2 + b^2
∣∣ ∣∣ ∣ =^
|ax 1 + by 1 + c| √ a^2 + b^2
Using the above formula the distance from (− 2 , 3) to the line 3x − 4 y + 5 = 0 is given by
√ 32 + (−4)^2
- (a) Find a point at which the given lines intersect:
r 1 (t) =< 1 , 1 , 0 > +t < − 1 , 1 , 2 >, t ∈ R r 2 (s) =< 2 , 0 , 2 > +s < − 1 , 1 , 0 >, s ∈ R
(b) Find the equation of the plane that contains these lines. Solution: (a) We can denote the lines as
r 1 (t) = 〈 1 − t, 1 + t, 2 t〉 , t ∈ R.
∣∣ ∣
P 1 P 0
∣∣ ∣ =
∣∣ ∣projα(β)
∣∣ ∣ =
∣∣ ∣∣ ∣
α.β |α|
∣∣ ∣∣ ∣
=
∣∣ ∣∣ ∣
λ(a(x 1 − x) + b(y 1 − y) + c(z 1 − z)) |λ|
a^2 + b^2 + c^2
∣∣ ∣∣ ∣ =^
|ax 1 + by 1 + cz 1 + d| √ a^2 + b^2 + c^2
- Find a point on the curve r(t) = 4 cos(t)i + 4 sin(t)j + 3tk at a distance 10π units from the origin along the curve in the direction of increasing arc length.
Solution: Since r(t) = 4 cos(t)i + 4 sin(t)j + 3tk is of the form r(t) = f (t)i + g(t)j + h(t)k, where f (t), g(t), h(t) are continuously differentiable functions in t, r′(t) = −4 sin(t)i + 4 cos(t)j + 3k and the arc length L(t 1 ) of r(t) from t = 0 to t = t 1 is given by: L(t 1 ) =
∫ (^) t 1
0
|r′(t)| dt =
∫ (^) t 1
0
√ 16(sin^2 (t) + cos^2 (t)) + 9 dt = 5t 1. If L(t 1 ) = 5t 1 = 10π, then t 1 = 2π, and the corresponding point on the curve is given by: r(2π) = 4 cos(2π)i + 4 sin(2π)j + 3(2π)k = 4i + 6πk.
- Reparametrize the curve
r(t) =
t^2 + 1
) i +
2 t t^2 + 1
j
with respect to the arc length measured from the point (1, 0) in the direction of increasing t. Express the parametrization in its simplest form. What can you con- clude about the curve?
Solution: Since r(t) =
t^2 + 1
) i +
2 t t^2 + 1
j,
the point (1, 0) corresponds to t = 0. Also f (t) =
t^2 + 1
− 1 and h(t) =
2 t t^2 + 1
are continuously differentiable at all t ∈ R.
r′(t) =
( − 4 t (t^2 + 1)^2
) i +
( − 2 t^2 + 2 (t^2 + 1)^2
) j, and
|r′(t)| =
√√ √√ ( − 4 t (t^2 + 1)^2
) 2
( − 2 t^2 + 2 (t^2 + 1)^2
) 2
t^2 + 1
Hence s(t) =
∫ (^) t
0
|r′(ξ)| dξ =
∫ (^) t
0
ξ^2 + 1
dξ = 2 tan−^1 (t)
⇒ t(s) = tan(
s 2
), for 0 ≤ s < π.
r(t(s)) =
( 2 (tan( 2 s ))^2 + 1
) i +
2 tan( s 2 ) (tan( s 2 ))^2 + 1
j = cos(s)i + sin(s)j 0 ≤ s < π.
Hence with this reparametrization, the points on the curve represent points on the upper half part of the unit circle centered at the origin, excluding the point (− 1 , 0).
- At what point does the curve y = ex^ have maximum curvature? What happens to the curvature as x → ∞?
Solution: κ(x) =
|y′′(x)| (1 + (y′(x))^2 )
3 2
ex (1 + e^2 x)
3 2
⇒ κ′(x) =
ex(1 − 2 e^2 x) (1 + e^2 x)
⇒ κ′(x) = 0 if and only if 1 − 2 e^2 x^ = 0 or x = −
ln(2).
Note that for x < −
ln(2), κ′(x) > 0 which implies
κ(x) is a strictly increasing function in (−∞, −
ln(2)).
For x > −
ln(2), κ′(x) < 0 which implies
κ(x) is a strictly decreasing function in (−
ln(2), ∞). Hence the point of the curve at which curvature is maximum is given by (−
ln(2), e−^
(^12) ln(2) ) = (−
ln(2),
Note that ex^ < 1 + e^2 x, for all x ∈ R, ⇒ 0 ≤ κ(x) =
ex (1 + e^2 x)
32 ≤^
√ (1 + e^2 x)
Since limx→∞
√ (1 + e^2 x)
= 0, ( 1 ) ⇒ (^) xlim→∞ κ(x) = 0.
- Find the unit tangent vector, unit normal vector and the binormal vector of the curves at the corresponding points given below:
(a) r(t) =
〈 t^2 ,
t^3 , t
〉 ,
( 1 ,
)
(b) r(t) = 〈cos(t), sin(t), ln(cos(t))〉 , (1, 0 , 0).
Solution: (a) r(t) =
〈 t^2 ,
t^3 , t
〉 ,
⇒ r′(t) =
〈 2 t, 2 t^2 , 1
〉 , hence the unit tangent vector at r(t) of the curve is given by:
⇒ T(t) =
r′(t) |r′(t)|
〈 2 t 2 t^2 + 1
2 t^2 2 t^2 + 1
2 t^2 + 1
〉 .
Since the point
( 1 ,
) of the curve r(t) corresponds to t = 1, the the unit tangent vector at that point is given by: T(1) =
〉 .
Also T′(t) =
〈 2 − 4 t^2 (2t^2 + 1)^2
4 t (2t^2 + 1)^2
− 4 t (2t^2 + 1)^2
〉 .
Since N(t) =
T′(t) |T′(t)|
, N(1) =
T′(1)
|T′(1)|
〈 −
〉 , and
B(1) = T(1) × N(1) =
〈 −
〉 .
(d) lim (x,y)→(0,0)
xy^3 x^2 + y^6
(e) lim (x,y)→(0,0)
x sin(x^2 + y^2 ) x^2 + y^2
(f) lim (x,y)→(4,π)
x^2 sin
( y x
)
(g) lim (x,y)→(0,1) f (x, y),
where f (x, y) = √xx+−y√− 11 −y if x + y 6 = 1 = 0 if x + y = 1.
Solution: (a) f (x, y) =
xy x^2 + y^2
m 1 + m^2
, for all (x, y) such that y = mx, x 6 = 0,
⇒ lim (x,y)→(0,0)
xy x^2 + y^2
m 1 + m^2
along y = mx,
which is different for different straight lines passing through the origin.
hence lim (x,y)→(0,0)
xy x^2 + y^2
does not exist.
( For a more detailed discussion refer to () below ). (() For all (x, y) on the line y = 0, f (x, y) = f (x, 0) = 0.
Similarly for all (x, y) on the line y = x, f (x, y) = f (x, x) =
⇒ |f (x, 0) − f (x, x)| =
for all x ∈ R. (1)
Take any ǫ ≤ 14 , say ǫ = 18. Since for all δ > 0 , however small, there exists points (x, y) of the straight line y = mx such that 0 <
x^2 + y^2 =
x^2 + m^2 x^2 < δ, (2) (1) and (2) implies that given ǫ = 18 , there exists no c ∈ R and no δ > 0,
such that |f (x, y) − c| < ǫ =
if 0 <
x^2 + y^2 < δ.)
(b) Clearly |x| ≤
√ x^2 + y^2 and |y| ≤
√ x^2 + y^2
⇒ |x| |y| ≤ (x^2 + y^2 )
⇒ |f (x, y)| =
|x| |y| √ x^2 + y^2
√ x^2 + y^2. (1)
Given ǫ > 0, take δ = ǫ, then from (1) it follows |f (x, y) − 0 | < ǫ if 0 <
x^2 + y^2 < δ.
⇒ lim (x,y)→(0,0)
xy √ x^2 + y^2
(c) Clearly
∣∣ ∣x^3
∣∣ ∣ = |x| x^2 ≤
√ x^2 + y^2 (x^2 + y^2 ) and
∣∣ ∣y^3
∣∣ ∣ = |y| y^2 ≤
√ x^2 + y^2 (x^2 + y^2 ),
|f (x, y)| =
∣∣ ∣∣ ∣
x^3 − y^3 x^2 + y^2
∣∣ ∣∣ ∣ ≤
∣∣ ∣∣ ∣
x^3 x^2 + y^2
∣∣ ∣∣ ∣ +
∣∣ ∣∣ ∣
y^3 x^2 + y^2
∣∣ ∣∣ ∣
≤ 2
x^2 + y^2 (x^2 + y^2 ) x^2 + y^2
√ x^2 + y^2. (1)
Given ǫ > 0 take δ = 2 ǫ then from (1) it follows |f (x, y) − 0 | < ǫ if 0 <
x^2 + y^2 < δ.
Hence lim (x,y)→(0,0)
x^3 − y^3 x^2 + y^2
(d) f (x, y) =
m 1 + m^2
, for all (x, y) such that x = my^3 , y 6 = 0. Hence by a similar argument as in part(a), lim (x,y)→(0,0)
xy^3 x^2 + y^6
does not exist.
(e) Note that lim (x,y)→(0,0)
x = 0, and lim (x,y)→(0,0)
sin(x^2 + y^2 ) x^2 + y^2
= lim r^2 → 0
sin r^2 r^2
where x = r cos(θ) and y = r sin(θ).
Hence lim (x,y)→(0,0)
x sin(x^2 + y^2 ) x^2 + y^2
( lim (x,y)→(0,0)
x
) ×
( lim (x,y)→(0,0)
sin(x^2 + y^2 ) x^2 + y^2
) = 0 × 1 = 0.
Aliter: Since
∣∣ ∣sin(x^2 + y^2 )
∣∣ ∣ ≤ (x^2 + y^2 ),
|f (x, y)| =
∣∣ ∣∣ ∣
x sin(x^2 + y^2 ) x^2 + y^2
∣∣ ∣∣ ∣ ≤ |x| ≤
√ x^2 + y^2. Given ǫ > 0, take δ = ǫ, then |f (x, y) − 0 | < ǫ if 0 <
x^2 + y^2 < δ.
(f) lim (x,y)→(4,π)
x^2 = lim x→ 4 x^2 = 16 and lim (x,y)→(4,π)
sin(
y x
) = lim u→ π 4
sin(u) = sin(
π 4
⇒ lim (x,y)→(4,π)
x^2 sin
( y x
)
( lim (x,y)→(4,π)
x^2
) ×
( lim (x,y)→(4,π)
sin(
y x
) = 16 sin(
π 4
(g) f (x, y) =
x + y − 1 √ x −
1 − y
if x + y − 1 6 = 0,
(x + y − 1)(
x +
1 − y) (
x −
1 − y)(
√ x +
1 − y)
x +
√ 1 − y.
Since lim (x,y)→(0,1)
x = lim x→ 0
x = 0 and lim (x,y)→(0,1)
√ 1 − y = lim y→ 1
√ 1 − y = 0, lim (x,y)→(0,1)
f (x, y) = 0, if x + y − 1 6 = 0. (1) Also f (x, y) = 0 if x + y − 1 = 0. (2) From (1) and (2) it follows, lim (x,y)→(0,1)
f (x, y) = 0.
- Examine whether
f (x, y) =
{ (^) xy(y (^2) −x (^2) ) x^2 +y^2 if^ (x, y)^6 = (0,^ 0) 0 if (x, y) = (0, 0)
is a continuous function.
Solution: Since f (x, y) is of the form f (x, y) =
g(x, y) h(x, y) where g, h are continuous functions and h(x, y) 6 = 0 for all (x, y) 6 = (0, 0), f (x, y) is continuous at all (x, y) 6 = (0, 0). To check the continuity of f at (x, y) = (0, 0).
|f (x, y)| =
|xy| |(y^2 − x^2 )| x^2 + y^2
|xy| (y^2 + x^2 ) x^2 + y^2
≤ |xy| ≤ (x^2 + y^2 ) for (x, y) 6 = (0, 0).
