Download 1. Find the tangent plane to the function: f(x, y) = sin(xy) at x = ,y = π

1. Find the tangent plane to the function:
f (x, y) = sin(xy)
at x = 12 , y =
π
2
2. Find the tangent plane to the function:
f (x, y) = 3x + 2y
at x = 17, y = 1
3. Find the volume of the parallelpiped spanned by the given 3
vectors
(a) (0, 0, 1); (0, 2, 0); (4, 0, 0)
(b) (1, 1, 2); (2, 3, 4); (−1, 2, 3)
4. Find the length of the curve
√ t
(x(t), y(t), z(t)) = (1 + et , 2 2e 2 , 3 + t)
0≤t≤5
5. Find the angle between the vectors v = (1, 2, 0) and w = (4, 8, 4)
6. (a) Find the projection of the line l(t) = (0, 0, 2) + t(0, 1, 0) onto the
plane x + y + z − 1 = 0.
(b) Find the projection of the line l(t) = (1, 1, 1) + t(1, 2, 4) onto the x-y
plane.
Solutions
1. Calculate the partial derivatives:
∂
f (x, y) = y cos(xy)
∂x
so
∂
f (x, y) = x cos(xy)
∂y
√
∂
1 π
π 2
f
,
=
∂x
2 2
4
√
∂
1 π
2
f
,
=
∂y
2 2
4
and the tangent plane is
√
√ √ 2 π 2
1
2
π
z=
+
x−
+
y−
2
4
2
4
2
2. The given function is just a plane. So the tangent plane is
z = 3x + 2y
1
3. (a) This parallelpiped will be a rectangular prism with height 1, length
2, and width 4. It’s volume is 8. (You can also find this as in the
next problem).
(b) We can find this by finding the determinant of the matrix


1 1 2
 2 3 4
−1 2 3
which comes out to be 5.
4. The speed of the curve is
p
x0 (t)2 + y 0 (t)2 + z 0 (t)2 = et + 1
(Note: e2t + 2et + 1 = (et + 1)2 ) Then calculate
Z
5
(et + 1)dt = e5 − 4
0
5.
v·w
20
5
= √ √ =√ √
|v||w|
4 5 6
5 6
So the angle is
−1
sin
5
√ √
5 6
6. (a) The normal to the plane is n = (1, 1, 1).
The projection of (0, 0, 2) onto the plane (x, y, z) · n + d = 0 is
(0, 0, 2) −
((0, 0, 2) · n + d)
|n|2
which comes out to be 13 (−1, −1, 5).
The projection of the vector (0, 1, 0) to be parallel to the plane is
(0, 1, 0) −
((0, 1, 0) · n)
1
n = (−1, 2, −1)
|n|2
3
So the projected line is
1
((−1, −1, 5) + t(−1, 2, −1))
3
(b) To project onto the x − y plane we can just drop the last coordinate,
so the answer is
l(t) = (1, 1, 0) + t(1, 2, 0)
2