STEYNING & DISTRICT U3A

Discovering Mathematics

Session 39

[Repeat of Session 3 with minor mods.]

Differential & Integral Calculus

preceded by some algebraic manipulation

& volumes of solids.

Solution of Quadratic Equations

The standard form of a quadratic equation is; ax2 + bx + c = 0

Derive the general solution; x = [-b +- √(b2 – 4ac)]/2a

Hint; Divide both sides of the standard equation by a & move c/a to right side.

Add b2/4ac to both sides, then take square root of both sides.

(ax2 + bx + c = 0)/a or (ax2 + bx)a = -c/a

x2 + bx/a = -c/a Adding b2/4ac to both sides gives.

x2 + bx/a + b2/4ac= b2/4ac - c/a

Factorize LHS; (x +b/2a)2 = b2/4ac - c/a = (b2 – 4ac)/4a2

Take square root of both sides; x + b/2a = [+-√(b2 – 4ac)]/2a

or; x = --b/2a +- √(b2 – 4ac)/2a = [-b +- √(b2 – 4ac)]/2a

Spherical & Conical Volumes

A conical hole is drilled into an inverted hemisphere of radius R. The cone has a base diameter of R/2 and its height is also R/2. Find the volume of material remaining after drilling. R = 2r & h = R/2

R r

h

Volume of hemisphere, Vs = 2πR3 /3 Volume of cone, Vc = πr2h/3 = π(R/2)2*(R/2)/3 = πR3/24 Therefore material remaining = Vs – Vc = 2πR3 /3 - πR3/24 Simplifying; [16πR3 - πR3]/24 = 15πR3/24 = 5πR3/8

Differential & Integral Calculus

Put very simply we may say that :

Differential Calculus consists of finding the value of very small elements,

And that

Integral Calculus is a means of summating the small elements into a whole.

Differential Calculus In a study of differential calculus, we will frequently come across expressions

such as dx, dy, dt & du etc.

We must regard these expressions to mean ‘minute bits’ of’ x, y, t & u etc.

The mathematical term for these expressions is ‘differentials’.

Because dx, say, is extremely small, the mathematical convention is to

regard dx times dx, ie (dx)2, to be of second order of smallness and therefore

negligible. Obviously, 3rd , 4th & higher orders of smallness can also be

ignored.

For example, let us assume that the function x increases by a minute amount

dx. We then have x+dx. If we square this we get (x+dx)2 = x2 + 2xdx + (dx)2

The 2nd term is of 1st order smallness and is significant. However (dx)2 is of

2nd order of smallness and can be ignored..

If we assume that dx is 1/100 of x, then (dx)2 is 1/10000 & is insignificent.,

whereas 2xdx = 2x/100 and is significent.

This principle can be well illustrated by a geometric representation.

x

x

dx

dx (dx)2 The areas total x2 + 2xdx + (dx)2

& one can see that the 3rd term is of little significance. If we cube the original expression, ie (x + dx)3, it will expand to x3 + 3x2dx + 3x(dx)2 +(dx)3

In this example the 3rd & 4th factors may be

ignored.

In differential calculus we are looking for the ratio of dy/dx for instances where

y and x are variables which are explicitly related to each other.

Assume y = x2. Then y+dy = (x + dx)2 = x2 + 2xdx + (dx)2

If we deduct the original equation from both sides & ignoring 2nd order items.

Then dy = 2xdx & dy/dx = 2x. This is the differential of y = x2

In a further example, y = x3 and y + dy = (x + dx)3

= x3 + 3x2dx + 3x(dx)2 +(dx)3

If we deduct the initial equation and ignore 2nd & 3rd order elements ;

dy = 3x2dx & dy/dx = 3x2

If we plot a curve of y= x3, the slope of the curve at any point x = dy/dx

For the general equation, y = axn the differential dy/dx = anx(n-1)

For an equation with a more complex relationship, the same principles apply.

Eg y= 3x4 + 2x3 – 5x2 + 2x + 5 dy/dx = 12x3 + 6x2 – 10x + 2

NB The constant 5 disappears on differentiating.

Try these

i) y = x4 – 7x2 + 5x + 15

dy/dx = 4x3 – 14x + 5

ii) u = 2v3 + 6v2 – 10v + 12

du/dv = 6v2 + 12v – 10

iii) s = 2/v3 + 6/v2 – 10/v = 2v-3 + 6v-2 – 10v-1

ds/dv = -6v-4 – 12v-3 + 10v-2 = -6/v4 – 12/v3 + 10/v2

Differentiating Products & Quotients

Products

For a relationship such as y = (x3 + 3)(5 – x2) we use a 2 step process.

Let (x3 + 3) = u & du/dx = 3x2 (5 – x2) = v & dv/dx = -2x

So y = uv & y+dy = (u+du)(v+dv) = uv + vdu + udv + dudv

As previously, deduct the original equation & discard 2nd order elements.

Then dy = v.du + u.dv & dy/dx = v.du/dx + u.dv/dx

So, dy/dx = (5 – x2)*3x2. + (x3 + 3)*-2x

Simplifying; dy/dx = 3x2(5 – x2). – 2x(x3 + 3) = 15x2 – 3x4 – 2x4 – 6x

& dy/dx = 15x2 – 5x4 – 6x

Quotients

For an equation such as ; y = (x3 + 3)/(5 – x2) we use a similar 2 step method

Ie y = u/v & y + dy = (u + du)/(v + dv)

Where u = (x3 + 3) and v = (5- x2)

The dividing process is somewhat more complicated than for a product, but after

discounting higher order elements;

y + dy = u/v + du/v – u.dv/v2 = u/v + (v.du – u.dv)/v2

After deducting the original equation & dividing both sides by dx

dy/dx = (v.du/dx – u.dv/dx)/v2

Replacing the values of u, v, du & dv, will produce;

dy/dx = [(5 – x2)(3x2) - (x3 + 3)(-2x)] /(5 – x2)2

dy/dx = (15.x2 - 3x4 + 2x4 + 6x) /(5 – x2)2

dy/dx = (15.x2 - x4 + 6x)/(5 – x2)2

Practical use of a Differential

A 10 m. ladder is placed with the foot 5m from the wall. If

the foot is moved a further 20 cm away from the wall, how

much lower on the wall will the head of the ladder be?

Using Pythagoras; l2 = h2 + x2 or h2 = l2 – x2.

Therefore; h = (l2 – x2)½.

Differentiating; dh/dx = ½ (l2 – x2)-½*(-2x)

Simplifying; dh = -x*dx/√ (l2 – x2)

but l = 10m, x = 5m & dx = 0.2m

and dh = -5*0.2/√(100 – 25) = -1/√75 = -√3/15

This is approx. -0.115m - but why negative?

h

x dx

dh

l

Integral Calculus

As stated earlier, Integral Calculus can be simply defined as a process for

summation of many very small elements.

Fundamentally the process of integration is the converse of differentiation.

Eg differential of y = axn is dy/dx = nax(n-1) or dy = nax(n-1).dx

The integral of the same equation would reverse that process thus :

y = ∫ (a.x(n).dx) = [(a.x(n+1))/(n+1) + c] where c is a constant.

The addition of a constant seems confusing but if you recall all constants

disappear when we differentiate, so we must include them to integrate.

Try these : y = ∫ (2x3dx) then y =[ x4/2 + c]

s = ∫ (2x2dt) then s = [2x3/3 + c]

Eg. The general integral of y = ∫ (bxn.dx) = [ bx(n+1)/(n + 1)+ c]

Let’s try a couple of practical uses of integration.

Volume of a Cone by Calculus

x

r

h

y

dy

Volume of thin slice, dV = πx2 times dy.

By similar triangles, x/y = r/h

ie x = ry/h & squaring both sides x2 = r2y2/h2

∴ dV = πr2y2/h2 times dy

By integration, add up all the slices from the

base to the peak of the cone,

ie from y=0 to y=h. Constant = c.

Volume V = πr2/h2 y2ℎ

0dy

V = πr2/h2 [y3/3 + c]0h , Substitute h & 0 for y

V= πr2h3/3h2 = πr2.h/3

For any conical body, V=Area of base*h/3

Volume of a Sphere by Integral Calculus

r

x

y

Volume of slice, dV = πx2 . dy

x2 + y2 = r2 or x2 = r2 - y2

& dV = π(r2 - y2 ) . dy

Integrate to summate all slices from y = 0 to y = r

V= π (𝑟

0r2 - y2 ) . dy = π.[(r2 y - y3/3 +c) ]0

r

Substitute r for y & also 0 for y.

V= π.[(r3– r3/3) +c] - π.[(r2*0 – 03/3) +c] = π. (3r3 - r3)/3

So V = 2πr3 /3 for a hemisphere.

For a sphere Volume = 4πr3 /3.

dy

Ideas for Future Topics ?