Given the displacement, to find the velocity, we differentiate. To find the acceleration, we differentiate twice.

Given the acceleration, to find the velocity, we integrate. To find the displacement, we integrate twice.

Example

A particle P moves on the- axis. At time t seconds, its acceleration isWhen its velocity isWhenthe displacement is 2m. Find expressions in terms of for the velocity and displacement.

When

When

Example:

The displacement of a particle isFind the velocity and acceleration in terms of t. When is the particle at rest for? What are the displacement and acceleration at this time?

When the particle is at rest,

The particle is at rest whenor

The displacement whenis

The acceleration whenis

{jatex options:inline}a= \frac{dv}{dt}=2v^2{/jatex}

Now separate variables.

{jatex options:inline}\frac{dv}{v^2}=v^{-2}dv=dt{/jatex}

Now integrate in the usual way.

{jatex options:inline}\int^v_1 v^{-2}dv = \int^t_2 dt{/jatex}

{jatex options:inline}[- \frac{1}{v} ]^v_1 = t-2{/jatex}

{jatex options:inline}- \frac{1}{v} - ( \frac{1}{1} )=t-2 {/jatex}

Making {jatex options:inline}v{/jatex} the subject gives {jatex options:inline}v= \frac{1}{3-t}{/jatex}.]]>

Suppose that a spherically shaped raindrop is falling. As it falls, water condenses on it. The rate of increase of the radius of the raindrop is 0.01 mm/s at a tuime when the radius of the raindrop is 3 mm. What is the rate at which the volume is increasing?

If the raindrop can be treated as a sphere, then the volume of the raindrop is The chain rule may be used in the form

so whenmm,

We are told in the question that the rate of increase ofmm/s, so

mm ^{3 }/s.

Suppose on the other hand that air is being blown into a balloon at the rate of 5 cm ^{3 }/s at a point when the radius of the balloon is 6 cm. Find the rate at which the radius of the ballon is increasing.

Again using the relationshipwithso when r=6 cm,

is the rate at which air is being blown into the balloon, so

cm/s.

]]>Suppose that a cube is increasing in volume at the rate of 4 cm ^{3 }/s. The cube has sides of lengthso the volume of the cube isand the surface area isSuppose it is desired to find the rate at which the surface area is increasing when

We can use the Chain Rule, which states

and

From the first of these, withso that whenthen

cm/s

From the second of these, withso that when

cm ^{2 } /s.

Alternatively, we could have used the Chain Rule in the form

With

Whensoand

cm ^{2 }/s.

and the marginal cost is the increase in the total cost as a result of producing one more of the product.

In practice, the marginal cost is taken to be the slope of the cost function at

A typical cost finction is shown below.

The total cost increases quite quickly at first, as fixed costs – costs regardless of the level of production – must be met. At a certain point, economies of scale become significant, and as production increases further, capacity constraints become significant, leading to a rapid increase in the cost function as more investment must be made to increase production further.

Of special interest to manufacturers is the point at which the average cost is miniminized. On the diagram below the average cost is the slope of the line from the origin to the point

The slope, and so the average cost, is a minimum when average cost equals marginal cost, as shown below.

We can find the output that minimises the average cost using the equation

Example: A factory has a cost functionFindthat minimises the average cost.

We solve

hence

]]>If the business sellsitems the total revenue would be

If the cost of producing theitems isthe profit is given by

The aim of the business is to maximise profit. Profit will be maximised when orThis means that profit is maximised when marginal revenue equals marginal cost. To ensure that profit is maximised – and not minimised – we should use the second derivative test. Ifthen profit is maxised.

If the cost functionand price functionthe profit isso and setting this equal to zero gives x=11.94.

so profit is maximised.

]]>We can then integrate both sides:

Example: Solve the differential equation

Multiply byand divide byto giveWe can now integrate:

is a product which we integrate by parts obtainingTo find the constantwe need what is called a boundary condition. Suppose then that we have that whenSubstitute these values into (1) to obtain

hence

Example: Solve

Factorise the right hand side intoto givewhich is separable.

Multiply byand divide byto giveWe can now integrate:(1)

If we are to makethe subject we exponentiate both sides, raisingto the power of both sides:

whereNotice how the constant termin (1) becomes the constant factorwhen we exponentiate.

]]>Givento findwe differentiateTo findwe differentiateWe can differentiate any number of times: if we differentiate y n times, we have

Examples of differential equations include:

If a functionis a solution to the differential equation then we can substitutefor into the equation and obtain an identity.

Example: Show thatsatisfies the equation

Hence the given function satisfies the equation.

Example: Show thatsatisfies the equation

Hence the given function satisfies the equation.

]]>The last expression is of the formwhich differentiates toApplying this example towe obtainDifferentiatingonly introduces another factorto giveWe can then find tangents and normals in the usual way.

Example: Find the equation of the tangent and normal to the curveat the point with coordinates

For the tangent

so at the point

For the normal

]]>The formula for the difference quotient for a functionat a pointis

Example: Find the difference quotient for the function

]]>