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Devansh Jain
IIT Bombay
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Sheet 6
2 – Directional Derivative in R2
4 – Directional Derivative in R3
5 – Mixed Partial Derivative
8 – Maxima, Minima and Saddle Points
9 – Absolute Maximum and Minimum
2. f (x,y) = x2 + sin(xy)
First, as fx = 2x + y cos(xy) and fy = x cos(xy) are continuous, we note that f (x,y) is differentiable at the point (1,0).
This implies that, for a unit vector u, we have Duf (1,0) = ((∇f )(1,0)) · u.
We have to find a u = (u1,u2) so that Duf (1,0) = 1.
Plugging in x=1 and y=0 in fx and fy , we get that (∇f )(1,0) = (2,1)
Thus, we want (2,1) · (u1,u2) = 1, that is 2u1 + u2 = 1 =⇒ 2u1 = 1 − u2. As u is a unit vector, we also have u

=⇒ u2 = 1 or u .
The corresponding unit vectors are u=(0,1) and u=(4/5,−3/5)
Indeed we see that u = (0,1) and u = (4/5,−3/5) do imply Duf (1,0) = 1.
Thus, these are the only directions where directional derivative takes the value 1.
Basically we have proved that Duf (1,0) = 1 ⇐⇒ u = (0,1) and u = (4/5,−3/5).

4. It is not too tough to show that the direction of the normal to a sphere at a pointon the sphere is the same as the direction of the vector joining the center to that point.
Indeed, we get that (∇S)(x0,y0,z0) = 2(x0,y0,z0), where S(x,y,z) := x2 + y2 + z2 for (x,y,z) ∈ R3.
Thus, the required u is .
Hence,
(D .
5. We shall assume that z is a “sufficiently smooth” function of x and y. We are given that sin(x + y) + sin(y + z) = 1 and cos(y + z) 6= 0. Differentiating with respect to x while keeping y constant gives us
cos(x + y) + cos(y + z)∂∂xz = 0. (∗)
Similarly, differentiating with respect to y while keeping x constant gives us
. (∗∗)
Differentiating ( y gives us

1 ∂2z ∂2z Note that I have implicitly assumed that ∂x∂y = ∂y∂x. However, using a different set of calculations, one can arrive at the same answer without assuming this. I encourage you to try that.
Thus, using (∗) and (∗∗), we get
∂
∂x
1
cos(
cos2(x + y)
= + tan(y + z)
cos2(y + z)

8. (i) .
Note that the above function is defined on D = R2.
Thus, every point is an interior point of D. Moreover, it can be seen that the partial derivatives of all orders exist and are continuous everywhere.
For (x0,y0) to be a point of extrema or a saddle point, it must be the case that (∇f )(x0,y0) = (0,0).
Note that fx . Also, fy .
Thus, solving (∇f )(x0√,y0) = (0,√0) gives us that√ √
(x0,y0) ∈ {(0,0), (0, 2), (0,− 2), (− 2,0), ( 2,0)}.
Now, we determine the exact nature using the determinant test.
Recall that (∆f )(x0,y0) := fxx (x0,y0)fyy (x0,y0) − fxy (x0,y0)2 .
Hence, in our case,

Moreover, fxx(x,y) = e−(x2+y2)/2(x4 − x2y2 − 5×2 + y2 + 2)
For (x0,y0) = (0,0), it is clear that it is a saddle point for f as discriminant is −4 < 0.
Note that if x = 0,√the discriminant reduces to −e−y2(y6 − 3y4 − 8y2 + 4).

Substituting y = ± 2 gives us that the discriminant is positive with fxx positive and hence, the points are points of local minima.
√
Similarly, we get that the points (± 2,0) are points of local maxima as they have discriminant positive and fxx negative.
8. (ii) f (x,y) = f (x,y) = x3 − 3xy2.
Note that the above function is defined on D = R2.
Thus, every point is an interior point of D. Moreover, it can be seen that the partial derivatives of all orders exist and are continuous everywhere.
For (x0,y0) to be a point of extrema or a saddle point, it must be the case that (∇f )(x0,y0) = (0,0).
Note that fx(x,y) = 3×2 − 3y2. Also, fy(x,y) = −6xy.
Thus, solving (∇f )(x0,y0) = (0,0) gives us that (x0,y0) = (0,0).
Now, we determine the exact nature using the determinant test.
Recall that (∆f )(x0,y0) := fxx (x0,y0)fyy (x0,y0) − fxy (x0,y0)2 . Hence, in our case,
.
Thus, for (x0,y0) = (0,0), we get the discriminant is 0.
Hence, we get that the discriminant test is inconclusive!
This means that we must turn to some other analytic methods of determining the nature.
Now, we note that f (δ,0) = δ3 for all δ ∈ R.
Thus, given any , choose .
This gives us that (0,0) is saddle point.
9. To find: Absolute maxima and minima of
f y for 1 ≤ x ≤ 3,−π/4 ≤ y ≤ π/4.
Note that the domain is a closed and bounded set. As f is continuous on the domain, f does achieve a maximum and a minimum. Note that fx(x,y) = (2x − 4)cosy and fy y for interior points (x,y).
Thus, the only critical point is p1 = (2,0).
Now we restrict ourselves to the boundaries to find the local extrema.
“Right boundary:” This is the line segment x = 3,−π/4 ≤ y ≤ π/4.
The function now reduces to −3cosy on this segment.
Using our theory from one-variable calculus, we get that we need to check the points (3,0), (3,π/4), (3,−π/4).
Similar consideration of the “left boundary” gives us the points (1,0), (1,π/4), (1,−π/4).

Now, we look at the “top boundary.”
The function there reduces to x2√− 4x.
2
Once again, using our theory from one-variable calculus, we get that we need to check the points (1,π/4), (2,π/4), (3,π/4).
Similarly, checking the “bottom boundary” gives us the points (1,−π/4), (2,−π/4), (3,−π/4).
We now tabulate our results as follows:
(x0,y0) (2,0) (3,0) (3,π/4) (2,π/4) (1,π/4)
f (x0,y0) −4 −3 −4 √
2
(x0,y0) (1,0) (1,−π/4) (2,−π/4) (3,−π/4)
f (x0,y0) −3 −3 √
2
Thus, we get that fmin = −4 at (2,0) and f at (1,±π/4) and (3,±π/4).
References