Quod Erat Demonstrandum

2009/09/25

Find dy/dx at a point not on the curve

Filed under: Additional / Applied Mathematics,HKCEE — johnmayhk @ 4:50 下午
Tags: , ,

When distributing the marked test paper to students, one student, Carman, reminded me that there was a ‘question’ in the following question:

If x^3 - 4x^2y + 3xy^2 - y^5 = 10, find \frac{dy}{dx} at the point (-2,1).

Carman said, ‘the point does NOT lie on the curve.’

Good observation! I had to say thank you to him. Although I’m not the setter, I should bear the responsibility of checking the paper.

But a natural follow-up question turns up: what is the meaning of the number \frac{dy}{dx}|_{(-2,1)} = \frac{31}{33} we are obtaining? Is the number meaningless or standing for something?

Let’s consider a simple example.

\frac{dy}{dx} = 3x^2

Now, we are free to put values of x to obtain certain number, e.g. \frac{dy}{dx}|_{x=2} = 3(2)^2 = 12. Well, what is the meaning of this “12″?

Actually, we can easily solve the differential equation \frac{dy}{dx} = 3x^2 (by integration, say), and obtain

y = x^3 + C,

where C is an arbitrary constant.

The equation y = x^3 + C is actually representing a family of curves.

Then, no matter which point we are considering, like (2,0), (2,10) or even (2,1997), we can find a ‘member’ in the family such that the point really lie on that member!

For (2,0), set (0) = (2)^3 + C \Rightarrow C = -8, hence (2,0) lies on y = x^3 - 8
For (2,10), set (10) = (2)^3 + C \Rightarrow C = 2, hence (2,10) lies on y = x^3 + 2
For (2,1997), set (1997) = (2)^3 + C \Rightarrow C = 1989, hence (2,10) lies on y = x^3 + 1989

So, when we are finding \frac{dy}{dx}|_{x=2}, we are actually finding the slope of tangent at a point (2,a) to the curve y = x^3 + (a - 8), see the meaning?

Another example, say

\frac{dy}{dx} = -\frac{x}{y}

Free to put values of x and y into the above, what does the answer stand for?

On solving the differential equation \frac{dy}{dx} = -\frac{x}{y}, yield

x^2 + y^2 = C

for any value of C.

Student, you could see that it stands for a family of curves (in this case, circles), centred at O.

Hence, if we ask to find the value \frac{dy}{dx}|_{P(2,3)} say, we are actually calculating the slope of tangent to a circle \mathfrak{C} which is centred at O with radius OP.

Get it?

Now, looking back to the original question,

x^3 - 4x^2y + 3xy^2 - y^5 = 10

implies

\frac{dy}{dx} = \frac{3x^2 - 8xy + 3y^2}{4x^2 - 6xy + 5y^4}

But, conversely

\frac{dy}{dx} = \frac{3x^2 - 8xy + 3y^2}{4x^2 - 6xy + 5y^4}

DOES NOT IMPLY

x^3 - 4x^2y + 3xy^2 - y^5 = 10

but, as you can imagine,

x^3 - 4x^2y + 3xy^2 - y^5 = C (where C is an arbitrary constant)

satisfies the differential equation

\frac{dy}{dx} = \frac{3x^2 - 8xy + 3y^2}{4x^2 - 6xy + 5y^4}

Now x^3 - 4x^2y + 3xy^2 - y^5 = C represents equation of family of curves.

It is easy to find a member in this family, such that it passes through (-2,1), just set

(-2)^3 - 4(-2)^2(1) + 3(-2)(1)^2 - (1)^5 = C \Rightarrow C = -31

Hence, finding the value of \frac{dy}{dx} at (-2,1) can be interpreted as finding the slope of tangent to the curve

x^3 - 4x^2y + 3xy^2 - y^5 = -31 at (-2,1).

[SBA]

Apart from x^3 - 4x^2y + 3xy^2 - y^5 = C, will there be other equation(s) satisfying

\frac{dy}{dx} = \frac{3x^2 - 8xy + 3y^2}{4x^2 - 6xy + 5y^4}?

3 則迴響 »

  1. 呀sir~係唔係冇呀?
    假此真係有第二式符合既話
    當係f(x),g(x)

    d(f(x))/dx = d(g(x))/dx
    by integration
    f(x)+C = g(x)+C’

    咁係唔係講左其實g(x)其實係等價於f(x)?

    迴響 由 Carmen — 2009/09/27 @ 11:10 上午 | 回覆

  2. 回樓上: 其中一個[f or g]係implicit function 點解可以寫做 y=f(x) or y=g(x)呢?

    迴響 由 whsvin — 2009/09/27 @ 8:45 下午 | 回覆

  3. 回樓上:冇留意到,唔好意思….

    迴響 由 Carmen — 2009/09/27 @ 10:01 下午 | 回覆


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