Snowplow's problem

Setting the problem Solving the problem

Snowplow’s problem

Raul Romero Martın

[email protected]

IES Gabriel CisnerosDepartamento de Matematicas

Setting the problem Solving the problem

Indice

1 Setting the problem

2 Solving the problem

Setting the problem Solving the problem

The problem

One day, it begins to snow...

At noon, a snowplow starts to move for doing its work.In the first hour the snowplow have moved two kilometers...In the second hour, there is more snow, it is more difficult to moveand so... it only moves one more kilometer.

The question is...

What time did it begin to snow?

Setting the problem Solving the problem

The problem

One day, it begins to snow...At noon, a snowplow starts to move for doing its work.

In the first hour the snowplow have moved two kilometers...In the second hour, there is more snow, it is more difficult to moveand so... it only moves one more kilometer.

The question is...

What time did it begin to snow?

Setting the problem Solving the problem

The problem

One day, it begins to snow...At noon, a snowplow starts to move for doing its work.In the first hour the snowplow have moved two kilometers...

In the second hour, there is more snow, it is more difficult to moveand so... it only moves one more kilometer.

The question is...

What time did it begin to snow?

Setting the problem Solving the problem

The problem

One day, it begins to snow...At noon, a snowplow starts to move for doing its work.In the first hour the snowplow have moved two kilometers...In the second hour, there is more snow, it is more difficult to moveand so...

it only moves one more kilometer.

The question is...

What time did it begin to snow?

Setting the problem Solving the problem

The problem

One day, it begins to snow...At noon, a snowplow starts to move for doing its work.In the first hour the snowplow have moved two kilometers...In the second hour, there is more snow, it is more difficult to moveand so... it only moves one more kilometer.

The question is...

What time did it begin to snow?

Setting the problem Solving the problem

The problem

One day, it begins to snow...At noon, a snowplow starts to move for doing its work.In the first hour the snowplow have moved two kilometers...In the second hour, there is more snow, it is more difficult to moveand so... it only moves one more kilometer.

The question is...

What time did it begin to snow?

Setting the problem Solving the problem

The problem

One day, it begins to snow...At noon, a snowplow starts to move for doing its work.In the first hour the snowplow have moved two kilometers...In the second hour, there is more snow, it is more difficult to moveand so... it only moves one more kilometer.

The question is...

What time did it begin to snow?

Setting the problem Solving the problem

Indice

1 Setting the problem

2 Solving the problem

Setting the problem Solving the problem

Let us set the following definitions:

t ≡ time since noon

t0 ≡ time since it began to snow until noon in hours

x(t) ≡ space traveled by the snowplow at time t

v(t) ≡ snowplow’s velocity at time t

h(t) ≡ snow’s height at time t

c ≡ snowplow’s velocity with 1 meter of snow

v =c

hv(t) =

A

t + t0

x ′(t) =A

t + t0=⇒ x(t) = A · log(t + t0) + B

Setting the problem Solving the problem

Let us set the following definitions:

t ≡ time since noon

t0 ≡ time since it began to snow until noon in hours

x(t) ≡ space traveled by the snowplow at time t

v(t) ≡ snowplow’s velocity at time t

h(t) ≡ snow’s height at time t

c ≡ snowplow’s velocity with 1 meter of snow

v =c

h

v(t) =A

t + t0

x ′(t) =A

t + t0=⇒ x(t) = A · log(t + t0) + B

Setting the problem Solving the problem

Let us set the following definitions:

t ≡ time since noon

t0 ≡ time since it began to snow until noon in hours

x(t) ≡ space traveled by the snowplow at time t

v(t) ≡ snowplow’s velocity at time t

h(t) ≡ snow’s height at time t

c ≡ snowplow’s velocity with 1 meter of snow

v =c

hv(t) =

A

t + t0

x ′(t) =A

t + t0=⇒ x(t) = A · log(t + t0) + B

Setting the problem Solving the problem

Let us set the following definitions:

t ≡ time since noon

t0 ≡ time since it began to snow until noon in hours

x(t) ≡ space traveled by the snowplow at time t

v(t) ≡ snowplow’s velocity at time t

h(t) ≡ snow’s height at time t

c ≡ snowplow’s velocity with 1 meter of snow

v =c

hv(t) =

A

t + t0

x ′(t) =A

t + t0

=⇒ x(t) = A · log(t + t0) + B

Setting the problem Solving the problem

Let us set the following definitions:

t ≡ time since noon

t0 ≡ time since it began to snow until noon in hours

x(t) ≡ space traveled by the snowplow at time t

v(t) ≡ snowplow’s velocity at time t

h(t) ≡ snow’s height at time t

c ≡ snowplow’s velocity with 1 meter of snow

v =c

hv(t) =

A

t + t0

x ′(t) =A

t + t0=⇒ x(t) = A · log(t + t0) + B

Setting the problem Solving the problem

x(t) = A · log(t + t0) + B

and we know x(0) = 0, x(1) = 2, x(2) = 3.

This leads to

0 = A · log t0 + B =⇒ B = −A · log t0.

2 = A · log

(1 + t0

t0

).

3 = A · log

(2 + t0

t0

).

From here we get,

3·log

(1 + t0

t0

)= 2·log

(2 + t0

t0

)=⇒ log

(1 + t0

t0

)3

= log

(2 + t0

t0

)2

And so, we arrive to the equation (1 + t0)3 = (2 + t0)2 · t0.

Setting the problem Solving the problem

x(t) = A · log(t + t0) + B

and we know x(0) = 0, x(1) = 2, x(2) = 3.This leads to

0 = A · log t0 + B =⇒ B = −A · log t0.

2 = A · log

(1 + t0

t0

).

3 = A · log

(2 + t0

t0

).

From here we get,

3·log

(1 + t0

t0

)= 2·log

(2 + t0

t0

)=⇒ log

(1 + t0

t0

)3

= log

(2 + t0

t0

)2

And so, we arrive to the equation (1 + t0)3 = (2 + t0)2 · t0.

Setting the problem Solving the problem

x(t) = A · log(t + t0) + B

and we know x(0) = 0, x(1) = 2, x(2) = 3.This leads to

0 = A · log t0 + B

=⇒ B = −A · log t0.

2 = A · log

(1 + t0

t0

).

3 = A · log

(2 + t0

t0

).

From here we get,

3·log

(1 + t0

t0

)= 2·log

(2 + t0

t0

)=⇒ log

(1 + t0

t0

)3

= log

(2 + t0

t0

)2

And so, we arrive to the equation (1 + t0)3 = (2 + t0)2 · t0.

Setting the problem Solving the problem

x(t) = A · log(t + t0) + B

and we know x(0) = 0, x(1) = 2, x(2) = 3.This leads to

0 = A · log t0 + B =⇒ B = −A · log t0.

2 = A · log

(1 + t0

t0

).

3 = A · log

(2 + t0

t0

).

From here we get,

3·log

(1 + t0

t0

)= 2·log

(2 + t0

t0

)=⇒ log

(1 + t0

t0

)3

= log

(2 + t0

t0

)2

And so, we arrive to the equation (1 + t0)3 = (2 + t0)2 · t0.

Setting the problem Solving the problem

x(t) = A · log(t + t0) + B

and we know x(0) = 0, x(1) = 2, x(2) = 3.This leads to

0 = A · log t0 + B =⇒ B = −A · log t0.

2 = A · log

(1 + t0

t0

).

3 = A · log

(2 + t0

t0

).

From here we get,

3·log

(1 + t0

t0

)= 2·log

(2 + t0

t0

)=⇒ log

(1 + t0

t0

)3

= log

(2 + t0

t0

)2

And so, we arrive to the equation (1 + t0)3 = (2 + t0)2 · t0.

Setting the problem Solving the problem

x(t) = A · log(t + t0) + B

and we know x(0) = 0, x(1) = 2, x(2) = 3.This leads to

0 = A · log t0 + B =⇒ B = −A · log t0.

2 = A · log

(1 + t0

t0

).

3 = A · log

(2 + t0

t0

).

From here we get,

3·log

(1 + t0

t0

)= 2·log

(2 + t0

t0

)=⇒ log

(1 + t0

t0

)3

= log

(2 + t0

t0

)2

And so, we arrive to the equation (1 + t0)3 = (2 + t0)2 · t0.

Setting the problem Solving the problem

x(t) = A · log(t + t0) + B

and we know x(0) = 0, x(1) = 2, x(2) = 3.This leads to

0 = A · log t0 + B =⇒ B = −A · log t0.

2 = A · log

(1 + t0

t0

).

3 = A · log

(2 + t0

t0

).

From here we get,

3·log

(1 + t0

t0

)= 2·log

(2 + t0

t0

)

=⇒ log

(1 + t0

t0

)3

= log

(2 + t0

t0

)2

And so, we arrive to the equation (1 + t0)3 = (2 + t0)2 · t0.

Setting the problem Solving the problem

x(t) = A · log(t + t0) + B

and we know x(0) = 0, x(1) = 2, x(2) = 3.This leads to

0 = A · log t0 + B =⇒ B = −A · log t0.

2 = A · log

(1 + t0

t0

).

3 = A · log

(2 + t0

t0

).

From here we get,

3·log

(1 + t0

t0

)= 2·log

(2 + t0

t0

)=⇒ log

(1 + t0

t0

)3

= log

(2 + t0

t0

)2

And so, we arrive to the equation (1 + t0)3 = (2 + t0)2 · t0.

Setting the problem Solving the problem

x(t) = A · log(t + t0) + B

and we know x(0) = 0, x(1) = 2, x(2) = 3.This leads to

0 = A · log t0 + B =⇒ B = −A · log t0.

2 = A · log

(1 + t0

t0

).

3 = A · log

(2 + t0

t0

).

From here we get,

3·log

(1 + t0

t0

)= 2·log

(2 + t0

t0

)=⇒ log

(1 + t0

t0

)3

= log

(2 + t0

t0

)2

And so, we arrive to the equation (1 + t0)3 = (2 + t0)2 · t0.

Setting the problem Solving the problem

(1 + t0)3 = (2 + t0)2 · t0

After some computation we obtain t0 =

√5− 1

2= 0, 61803 hours.

So, t0 = 37 minutes 5 seconds. And our answer to the problem is...

It began to snow at 11:22:55

Setting the problem Solving the problem

(1 + t0)3 = (2 + t0)2 · t0

After some computation we obtain t0 =

√5− 1

2= 0, 61803 hours.

So, t0 = 37 minutes 5 seconds. And our answer to the problem is...

It began to snow at 11:22:55

Setting the problem Solving the problem

(1 + t0)3 = (2 + t0)2 · t0

After some computation we obtain t0 =

√5− 1

2= 0, 61803 hours.

So, t0 = 37 minutes 5 seconds. And our answer to the problem is...

It began to snow at 11:22:55

Setting the problem Solving the problem

(1 + t0)3 = (2 + t0)2 · t0

After some computation we obtain t0 =

√5− 1

2= 0, 61803 hours.

So, t0 = 37 minutes 5 seconds. And our answer to the problem is...

It began to snow at 11:22:55