MA1132 (Advanced Calculus) Tutorial sheet 4
[February 12 – 13 , 2013]
1. Find parametric equations for the tangent line to the parametric curve
x(t) = (t cosh t, t sinh t, t)
at the point where t = 0.
Solution: We need x0 (0) (which is a vector parallel to the tangent line) and x(0) = (0, 0, 0)
(which is a point on the line).
dx
= (cosh t + t sinh t, sinh t + t cosh t, 1)
dt
x0 (t) =
and so
x0 (0) = (1 + 0, 0 + 0, 1) = (1, 0, 1).
The line has parametric equations

 x = 0 + 1t
y = 0 + 0t

z = 0 + 1t
or

 x = t
y = 0

z = t
2. Find the partial derivatives with respect to x and y evaluated at (x0 , y0 ) = (2, −2) for
f (x, y) =
x cos(πy)
x2 + y 2
Solution: We could do this
f (x, y0 ) = f (x, −2)
x cos(−2π)
=
x2 + (−2)2
x
= 2
x +4
∂f
d
d
x
|(x,−2) =
f (x, y0 ) =
2
∂x
dx
dx x + 4
2
(x + 4) − x(2x)
=
(x2 + 4)2
4 − x2
=
(x2 + 4)2
∂f
|(2,−2) = 0
∂x
and this
f (x0 , y) = f (2, y)
2 cos(πy)
=
4 + y2
d
d 2 cos(πy)
∂f
|(2,y) =
f (x0 , y) =
∂y
dy
dy 4 + y 2
−2π sin(πy)(4 + y 2 ) − 2 cos(πy)(2y)
=
(x2 + 4)2
∂f
0+8
1
|(2,−2) =
=
2
∂x
8
8
But this is how we would usually do it:
cos(πy)(x2 + y 2 ) − x cos(πy)(2x)
∂f
=
∂x
(x2 + y 2 )2
(x2 + y 2 − 2x2 ) cos(πy)
=
(x2 + y 2 )2
2
(y − x2 ) cos(πy)
=
(x2 + y 2 )2
∂f
−πx sin(πy)(x2 + y 2 ) − x cos(πy)(2y)
=
∂y
(x2 + y 2 )2
−πx(x2 + y 2 ) sin(πy) − 2xy cos(πy)
=
(x2 + y 2 )2
∂f
|(2,−2) = 0
∂x
∂f
0+8
1
|(2,−2) =
=
2
∂y
8
8
∂f
3. Find the partials ∂f
and ∂f
evaluated at (x0 , y0 ) (which means
|(x ,y ) = fx (x0 , y0 ) and
∂x
∂y
∂x 0 0
∂f
|(x ,y ) = fy (x0 , y0 )) for the following cases.
∂y 0 0
(a) f (x, y) = e(x+2y−3) , (x0 , y0 ) = (−1, 1)
2
Solution:
∂f
=
∂x
=
∂f
=
∂y
=
e(x+2y−3)
∂
(x + 2y − 3)
∂x
e(x+2y−3)
e(x+2y−3)
∂
(x + 2y − 3)
∂y
2e(x+2y−3)
∂f
|(−1,1) = e−2
∂x
∂f
|(−1,1) = 2e−2
∂y
(b) f (x, y) =
Solution:
p
x2 + y 2 , (x0 , y0 ) = (−3, 4)
∂f
∂x
∂f
∂y
∂f
1
(2x)
= p
∂x
2 x2 + y 2
x
= p
x2 + y 2
∂f
1
= p
(2y)
∂y
2 x2 + y 2
y
= p
x2 + y 2
−3
|(−3,4) =
5
4
|(−3,4) =
5
3