parabola's equation

[psi20]

How do you solve this problem?

The parabola y=ax^2 + 6 is tangent to the line y=x.

Find a.

[Primarygun]

There is only one intersection for the two graphs.

[DQW]

How do you solve this problem?

The parabola y=ax^2 + 6 is tangent to the line y=x.

Find a.

A necessary condition for a line to be a tangent at some point of a curve is that the line have the same slope as the curve does at that point.

[psi20]

is a 1/2? I'm kind of rusty on derivatives.

[DQW]

Alternative approach : For a line to be tangent to a curve, the line and curve must have exactly one point of intersection. The points of intersection between the two is clearly given by the quadratic equation got from eliminating y between the two equations. What is the condition for a quadratic equation to have exactly one real solution ?

 

PS : 'a' is not 1/2

[DQW]

First approach :

 

Use the condition for equal slopes. This gives 2ax = 1 or x = 1/2a. From this, you have the point of intersection in terms of 'a', which you can plug into the equation for the parabola and solve for 'a'.

[ElijahJones]

How do you solve this problem?

The parabola y=ax^2 + 6 is tangent to the line y=x.

Find a.

 

Yeah I think it is easiest to simply realize that you are evaluating a constraint (sort of like a boundary condition)

 

f(x)=ax^2+6

g(x)=x

 

Such that f'(x)=g'(x), so

 

2ax = 1, x = 1/(2a)

 

This is the information we gathered from evaluating the constraint. But look we are really free to pick either a or x to suit our fancy. Nowhere does the problem stipulate that f(x) should equal g(x) only that at some point the slopes are equal.

 

So lets say we want a parabola that has slope one at the point x=1 then we get

 

1=1/2a, a = 1/2

 

So yes (a,x) = (1/2,1) is one solution, but for every x there is a corresponding a. In fact this is really a rather clever question. You can see that the real solution is a hyperbola by graphing x = 1/(2a). You could also think of this as a map that associates to every value x, a function f(x)=ax^2+6

 

Oh yeah also look at the parabola ax^2 + 6. The slope at any point is not affected by adding six (we know this from the derivative) and the parabola is still symmetric with respect to the y-axis. So if you want a parabola that has a slope of 1 at x=-1 pick a =-1/2. The reason this is interesting is that it flips the parabola upside down. This type of change with respect to a parameter is called a bifurcation, meaning the object is quantitatively different depending on whether a certain parameter (in this case x) has crossed a certain threshold (in this case zero). In more complicated systems of differential equations important properties of the system can change dramatically as parameters vary. I know that is far beyond the level of this question but maybe it will peak your interest.

 

[jcarlson]

Yeah I think it is easiest to simply realize that you are evaluating a constraint (sort of like a boundary condition)

 

f(x)=ax^2+6

g(x)=x

 

Such that f'(x)=g'(x)' date=' so

 

2ax = 1, x = 1/(2a)

 

This is the information we gathered from evaluating the constraint. But look we are really free to pick either a or x to suit our fancy. Nowhere does the problem stipulate that f(x) should equal g(x) only that at some point the slopes are equal.

 

So lets say we want a parabola that has slope one at the point x=1 then we get

 

1=1/2a, a = 1/2

 

So yes (a,x) = (1/2,1) is one solution, but for every x there is a corresponding a. In fact this is really a rather clever question. You can see that the real solution is a hyperbola by graphing x = 1/(2a). You could also think of this as a map that associates to every value x, a function f(x)=ax^2+6

 

Oh yeah also look at the parabola ax^2 + 6. The slope at any point is not affected by adding six (we know this from the derivative) and the parabola is still symmetric with respect to the y-axis. So if you want a parabola that has a slope of 1 at x=-1 pick a =-1/2. The reason this is interesting is that it flips the parabola upside down. This type of change with respect to a parameter is called a bifurcation, meaning the object is quantitatively different depending on whether a certain parameter (in this case x) has crossed a certain threshold (in this case zero). In more complicated systems of differential equations important properties of the system can change dramatically as parameters vary. I know that is far beyond the level of this question but maybe it will peak your interest.

 

[/quote']

 

This is incorrect, there is only one solution for a. You have to remember that in order for a function f(x) to be tangent to a function g(x), not only must f'(x) = g'(x) but also f(x) = g(x), or else you could end up with two lines that have the same slope, but dont touch (basically parallel at x).

 

if f(x) = ax^2 + 6 and g(x) = x, then we have the equations

 

ax^2 + 6=x AND

 

2ax = 1

 

Solving for x in the second equation, we have:

 

x= 1/(2a)

 

Now, solving the system by subsituting the second equation into the first, we have

 

a/(4a^2) + 6 = 1/(2a)

 

which simplifies to

 

6 = 1/2 * 1/a - 1/4 * 1/a

 

6= 1/(4a)

 

1/6 = 4a

 

a=1/24

 

Now, we check our work buy subbing a into the original equation:

 

f(x)=1/24 * x^2 + 6

 

f'(x) = 1/12 * x

 

g(x) = x

 

g'(x) = 1

 

1 = 1/12 * x

 

x = 12

 

At x = 12, we see that f(x) = 144/24 + 6 = 12, and g(x) = 12, therefore f(x) and g(x) intersect at this point, and f'(x) = 1 and g'(x) = 1, so they are in fact, tangent at x = 12, or when a = 1/24.