Review Questions

Question 1:

Findf_x,f_y andf_z if

f =zcos(y) +x

Answer:

f_x= 1 f_y=−zsin(y)

f_z= cos(y)

Question 2:

Findf_x if

f =xsin(xy) Answer:

fx= sin(xy) +xycos(xy)

Question 3:

Findfxx,fyy andfxy if

f =xsin(y) +yx²

Answer:

f_xx= 2y f_yy =−xsin(y) fxy = cos(y) + 2x

Question 4:

Find ∂z

∂x if the equation

yz−ln(z) =x+y definesz as a function ofxandy.

Hint: You will need to use implicit partial differentiation in this question. Refer to the section on implicit differentiation in the previous chapter if you need to.

Answer:

∂z

∂x = z yz−1

Question 5:

Using the chain rule, find ∂z

∂t if z= 2xy−y² x=t²+ 1, y=t²−1 and express the result in terms oft.

Medium Questions Question 6:

Find∇f if

f = y²+ sin(z) e^−x Answer:

∇f =− y²+ sin(z)

e^−x~i+ 2ye^−x~j+e^−xcos(z)~k

Question 7:

Let

f(x, y, z) =x²z+yz³

Find the directional derivative off at the point (1,3,2) in the direction to the point (2,1,3).

Answer:

D_uf(1,3,2) = 25

√6

Question 8:

Calculate div ~F if

F~ = (x²y,−xze^2y, x²y²)

Answer:

div ~F = 2xy−2xze^2y

Question 9:

Calculate curl ~F if

F~ = (xyz, x²cos(z), e^xy)

Answer:

curl ~F = (xe^xy+x²sin(z))~i−(ye^xy−xy)~j+ (2xcos(z)−xz)~k

Question 10:

Find∇²f if

f = 3x³y²z³ Answer:

∇²f = 18xy²z³+ 6x³z³+ 18x³y²z

Hard Questions

Question 11:

Using the chain rule, find ∂w

∂t if w=e^xyz

x=ts, y=s², z=t−s

Hint: You may wish to calculate all the partial derivatives you will need and substitute them into the formula for the chain rule.

Question 12:

By considering the function

f =x²yz+e^xyz show thatdiv(∇f) =∇²f.

Question 13:

Recall the relations between Cartesian coordinates and polar coordinates:

x=rcos(θ) andy=rsin(θ) Findr_x andθ_x.

Hint: Use implicit partial differentiation on two formulaer²=x²+y² and tan(θ) =x y. Answer:

rx= cos(θ) θx=−sin(θ)

r

Question 14:

Considerf g(x, y), h(x, y)

, wherex=x(t) andy=y(t, s). Findf_tandf_s. Hint: You will need to use the chain rule multiple times.

Answer:

f_t= ∂f

∂g

∂g

∂x dx dt +∂g

∂y

∂y

∂t

! +∂f

∂h

∂h

∂x dx

dt +∂h

∂y

∂y

∂t

!

fs= ∂f

∂g

∂g

∂y

∂y

∂s+∂f

∂h

∂h

∂y

∂y

∂s

Question 15:

Use Lagrange multipliers to find the maximum and minimum values of the function

f(x, y) =x²y subject to the constraint

g(x, y) =x²+y²−3 = 0

Answer:

fmin=−2 and fmax= 2

8 Integration

8.1 Introduction to Integration

Integration is the opposite of differentiation so it helps us find the answer to the question:

‘Suppose we have dy

dx = 2xthen what isy?’

From the previous section we know the answer isy=x²+C, whereCis a constant. So we say that the integralof 2xisx²+C.

Note: We have an infinite number of solutions to the above question as the constant C could be any number!

Graphically, we know that differentiation gives us the gradient of a curve. Integration, on the other hand, finds the area beneath a curve.

y

a b x

Suppose we wish to find the area under the curve between thexvaluesaandbin the graph above. Then we could find an approximate answer by creating the dashed rectangles above and adding their areas.

Now, we can see this does not give a perfect answer, however the smaller the width of the rectangles the smaller the error in our answer would be.

So if we could use rectangles of an infinitely small width we would find an answer with no error. This is what integration does. It creates an infinite number of rectangles under the curve all with an infinitely small width and sums their area to find the total area under the curve.

−2. −1. 1.

−2.

−1.

1.

2

0

This means that if we are integrating to find the area shaded in blue above, we will need to separate our calculation into two parts: one to calculate the area under the x-axis and one to calculate the area above thex-axis. We would then have to add up the modulus of our two areas to get the total positive area.

Notation

We use the fact that the integral of 2xisx²+Cto help introduce the following notation for integration.

If we want to integrate 2xwe write this as:

Z

2x dx=x²+C or Z

dx2x=x²+C

ˆ The Z

tells us we need to integrate.

ˆ The dx tells use to integrate with respect to x. So we could write Z

2z dz =z²+C but in this case we have integrated with respect toz.

ˆ We call Z

2x dx=x²+C the integral.

ˆ We call the expression that we are integrating the integrand. So in this case the integrand is 2x.

Another notation which we can use is the following:

If we have the functionf(x) then we write the integral of this asF(x). So iff(x) = 2xthenF(x) =x²+C.

Integration finds the area under the curve. So if we want to find the area between two points on the x-axis, sayaandb, we use the following notation:

Z b a

f(x)dx

This tell us we want to find the integral of f(x) and then use that information to find the area under the curve between aandb. We callaand b the ‘limits’ of this integral. We find the area by using the formula below:

Z b

a

f(x)dx=F(b)−F(a) Note: This is explained in detail in the section on definite integrals.

ˆ When we are given limits then we have a definite integral.

ˆ When we are not given limits then we have an indefinite integral.

Rules for Integrals

Integration is alinearoperation. This means that:

1. We can split up a sum into separate integrals. For example Z

(x³+ 2x+ 7)dx = Z

x³dx+ Z

2x dx+ Z

7dx. Note that each of these integrals will result in a constant, however since these are all arbitrary numbers, we can combine them into a single constantC.

2. We can take constants out of the integral. For example Z

6x⁴dx = 6 Z

x⁴dx. Note that the

8.2 Standard Integrals