Math 212 Spring 2008: Solutions: HW #3
Instructor: S. Cautis
1. section 2.3, #1bd
1b) ∂f /∂x = yexy and ∂f /∂y = xexy
2
1d) ∂f /∂x = (2x log(x2 + y 2 )) + (x2 + y 2 ) x22x
+y 2 and ∂f /∂y = (2y log(x +
2y
y 2 )) + (x2 + y 2 ) x2 +y2 . If you assumed log was not in base e then you need
to divide the second summands by log(e).
2. section 2.3, #6bd
The tangent planes at the point (0, 1) for the two functions in exercise 1b
and 1d are:
z − f (0, 1) = ∂f /∂x(0, 1)(x − 0) + ∂f /∂y(0, 1)(y − 1)
Substituting we get
z−1=x
and
z = 2(y − 1)
3. section 2.3, #8ab
8a)
ex
0
y cos(xy) x cos(xy)
8b)
1
0
−1 0
1 1
4. section 2.3, #13
13a)
∇f = (exp(−x2 −y 2 −z 2 )−2x2 exp(−x2 −y 2 −z 2 ), −2xy exp(−x2 −y 2 −z 2 ), −2xz exp(−x2 −y 2 −z 2 ))
13b and 13c are computed similarly.
5. section 2.3, #14
14a) f (1, 0, 1) = e−2 so the tangent plane is
w − e−2 = (e−2 − 2e−2 )(x − 1) + 0(y − 0) − 2e−2 (z − 1)
which simplifies to
w − e−2 = −e−2 (x − 1) − 2e−2 (z − 1)
14b and 14c are computed similarly.
6. section 2.4, #2
Since (x, y) = (sin t, cos t) describes a circle (x, y) = (2 sin t, 4 cos t) will describe an ellipse. Another way of seeing this is that 4x2 + y 2 = 4(4 sin2 t)+
(16 cos2 t) = 16. So we get an ellipse centred at the origin, with axis the
x and y axis and passing through the four points (±2, 0) and (0, ±4).
7. section 2.4, #10
The tangent vector when t = t0 is c′ (t0 ) = (6t0 , t20 ).
8. section 2.4, #12
The tangent vector when t = t0 is c′ (t0 ) = (2t0 , 0).
9. section 2.4, #16
When t = 0 the path passes through (1, 0, 0). The tangent direction is
given by the derivative (−2 cos t sin t, 3 − 3t2 , 1) evaluated at t = 0 which
equals (0, 3, 1). So the tangent line is
(1, 0, 0) + s(0, 3, 1) = (1, 3s, s)
where s is a parameter.
10. section 2.4, #18
c(0) = (e, e−1 , cos 1) while c′ (t) = (et , −e−t , − sin t) which equals (e, −e−1 , − sin 1)
when t = 1. So the position of the particle at time s after it flies off is
(e, e−1 , cos 1) + s(e, −e−1 , − sin 1)
We want the position when t2 = 2 which means s = t1 − t0 = 1. This
gives
(e, e−1 , cos 1) + (e, −e−1 , − sin 1) = (2e, 0, cos 1 − sin 1)
as the particle’s position at time t1 = 2.