## Description

Devansh Jain

IIT Bombay

Questions to be Discussed

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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) = ((โf )(1,0)) ยท 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 (โf )(1,0) = (2,1)

Thus, we want (2,1) ยท (u1,u2) = 1, that is 2u1 + u2 = 1 =โ 2u1 = 1 โ u2. As u is a unit vector, we also have u

=โ u2 = 1 or u .

The corresponding unit vectors are u=(0,1) and u=(4/5,โ3/5)

Indeed we see that u = (0,1) and u = (4/5,โ3/5) 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 โโ u = (0,1) and u = (4/5,โ3/5).

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 (โS)(x0,y0,z0) = 2(x0,y0,z0), where S(x,y,z) := x2 + y2 + z2 for (x,y,z) โ R3.

Thus, the required u is .

Hence,

(D .

5. We shall assume that z is a โsufficiently smoothโ 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)โโxz = 0. (โ)

Similarly, differentiating with respect to y while keeping x constant gives us

. (โโ)

Differentiating ( y gives us

1 โ2z โ2z Note that I have implicitly assumed that โxโy = โyโx. 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 (โ) and (โโ), we get

โ

โx

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 (โf )(x0,y0) = (0,0).

Note that fx . Also, fy .

Thus, solving (โf )(x0โ,y0) = (0,โ0) gives us thatโ โ

(x0,y0) โ {(0,0), (0, 2), (0,โ 2), (โ 2,0), ( 2,0)}.

Now, we determine the exact nature using the determinant test.

Recall that (โf )(x0,y0) := fxx (x0,y0)fyy (x0,y0) โ fxy (x0,y0)2 .

Hence, in our case,

Moreover, fxx(x,y) = eโ(x2+y2)/2(x4 โ x2y2 โ 5×2 + y2 + 2)

For (x0,y0) = (0,0), it is clear that it is a saddle point for f as discriminant is โ4 < 0.

Note that if x = 0,โthe discriminant reduces to โeโy2(y6 โ 3y4 โ 8y2 + 4).

Substituting y = ยฑ 2 gives us that the discriminant is positive with fxx positive and hence, the points are points of local minima.

โ

Similarly, we get that the points (ยฑ 2,0) are points of local maxima as they have discriminant positive and fxx negative.

8. (ii) f (x,y) = f (x,y) = x3 โ 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 (โf )(x0,y0) = (0,0).

Note that fx(x,y) = 3×2 โ 3y2. Also, fy(x,y) = โ6xy.

Thus, solving (โf )(x0,y0) = (0,0) gives us that (x0,y0) = (0,0).

Now, we determine the exact nature using the determinant test.

Recall that (โf )(x0,y0) := fxx (x0,y0)fyy (x0,y0) โ 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 (ฮด,0) = ฮด3 for all ฮด โ 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 โค x โค 3,โฯ/4 โค y โค ฯ/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 โ 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.

โRight boundary:โ This is the line segment x = 3,โฯ/4 โค y โค ฯ/4.

The function now reduces to โ3cosy on this segment.

Using our theory from one-variable calculus, we get that we need to check the points (3,0), (3,ฯ/4), (3,โฯ/4).

Similar consideration of the โleft boundaryโ gives us the points (1,0), (1,ฯ/4), (1,โฯ/4).

Now, we look at the โtop boundary.โ

The function there reduces to x2โโ 4x.

2

Once again, using our theory from one-variable calculus, we get that we need to check the points (1,ฯ/4), (2,ฯ/4), (3,ฯ/4).

Similarly, checking the โbottom boundaryโ gives us the points (1,โฯ/4), (2,โฯ/4), (3,โฯ/4).

We now tabulate our results as follows:

(x0,y0) (2,0) (3,0) (3,ฯ/4) (2,ฯ/4) (1,ฯ/4)

f (x0,y0) โ4 โ3 โ4 โ

2

(x0,y0) (1,0) (1,โฯ/4) (2,โฯ/4) (3,โฯ/4)

f (x0,y0) โ3 โ3 โ

2

Thus, we get that fmin = โ4 at (2,0) and f at (1,ยฑฯ/4) and (3,ยฑฯ/4).

References

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