Practice Problems: Limits, partial differentiation, linear approximation
1) Find the following limit:
x2 + y 2
lim
(x,y)→(0,0)
p
x2 + y 2 + 1 − 1
Answer: Converting to polar coordinates we obtain
lim √
r→0
r2
r2 + 1 − 1
Now we have a limit in one variable where the top and bottom go to zero, so we can use
L’Hopital’s rule.
√
2r
= lim 2 r2 + 1 = 2
lim √
r→0
r→0 r/ r 2 + 1
2) Find the first partial derivatives of w = ln(x + 2y + 3z).
Answer:
∂w/∂x = 1/(x + 2y + 3z)
∂w/∂y = 2/(x + 2y + 3z)
∂w/∂z = 3/(x + 2y + 3z)
3) Use implicit differentiation to find ∂z/∂x and ∂z/∂y where x2 + y 2 + z 2 = 3xyz. (This
means differentiate every term on both sides, treating x and y as independent variables but
z as a function of both of them. This will get an equation with x, y, z and zx or zy terms
depending on whether you differentiate with respect to x or y. Then solve for zx or zy .)
Answer: differentiating with respect to x we find
2x + 2zzx = 3yz + 3xyzx =⇒ zx (3xy − 2z) = 2x − 3yz
2x − 3yz
=⇒ zx =
3xy − 2z
By symmetry
zy =
2y − 3xz
3xy − 2z
4) Find an equation of the tangent plane at (1, 1, 2) to the surface
f (x, y) = z = 3y 2 − 2x2 + x
Answer: fx (x, y) = −4x + 1 =⇒ fx (1, 1) = −3. Similarly fy (x, y) = 6y =⇒ fy (1, 1) = 6.
So using z − z0 = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) with (x0 , y0 , z0 ) = (1, 1, 2) gives
z − 2 = −3(x − 1) + 6(y − 1) =⇒ 3x − 6y + z = −1
5) Verify that f (x, y) = ex cos(xy) is differentiable at (0, 0) and find the linearization of f (x, y)
at (0, 0). Then verify Clairaut’s theorem holds for all (x, y).
Answer: Both ex and cos(xy) have all order derivatives everywhere in both variables, so their
product does too.
Linearization: fx (x, y) = ex cos(xy) − ex y sin(xy) and fy (x, y) = −xex sin(xy). So plugging in
(0, 0) we get fx (0, 0) = 1 and fy (0, 0) = 0. Also z0 = f (0, 0) = 1. So the linearization is:
z − 1 = 1(x − 0) + 0(y − 0) =⇒ z = x + 1
To verify Clairaut’s theorem: fxy = −xex sin(xy) − ex sin(xy) − ex xy cos(xy) = fyx .
6) Use linear approximation to estimate the amount of tin in a closed cylindrical tin can with
radius 4 cm and height 12 cm, if the tin is 0.04 cm thick on all sides. [I.e. you want to find a
change ∆V in volume because this will give the tin in the walls of the can.]
Answer: Volume is V = (base)(height) = πr2 · h. We know
∆V ≈ Vr ∆r + Vh ∆h = 2πr · h∆r + πr2 ∆h
We have (r0 , h0 ) = (4, 12) and we want to find the change V (4, 12) − V (4 − .04, 12 − .08). Note
that when finding the height we have .04 cm both on the top and the bottom.
V (3.96, 11.02) ≈ V (4, 12) + Vr (4, 12)(3.96 − 4) + Vh (4, 12)(11.02 − 12)
= 192π + (96π)(−0.04) + (16π)(−0.08)
= 192π − 5.12π
Thus the amount of tin is the volume of the can with the outer dimensions, minus the volume
of the can with the inner dimensions, so approximately
192π − V (3.96, 11.02) = 5.12π ≈ 16 cm3
Some general comments: the linearization (or linear approximation) of a function at a point
is the tangent plane to the graph of the function at that point. This will be an approximation
to the function f (x, y). I.e. the tangent plane can be written
z = z0 + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )
and this is an approximation in the sense that f (x, y) is approximately z0 + fx (x0 , y0 )(x −
x0 ) + fy (x0 , y0 )(y − y0 ) when (x, y) is close to (x0 , y0 ).
Estimate fx and fy at (1, 1).
Answer: to compute fx , we fix y and see how f changes moving only in the x-direction. We
see that from x = 0 to x ≈ 1.7, f changes from f = 0 to f = 2 (moving along the horizontal
2−0
= 20
,
line y = 1. So we take the change in f and divide by the change in x to get fx ≈ 1.7−1
7
ie. almost 3.
On the other hand, fy ≈
f (1,1.5)−f (1,1)
1.5−1
=
1−0
0.5
= 2.