# 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
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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 | 回應

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 | 回應

3. 回樓上:冇留意到,唔好意思….

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