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Apologies, but I'm confused.

Let $y = f(x)$ be curve such that at a point $(a,b)$ lying on it, the slope of the tangent kissing that point is $f'(x)$

Now, the equation of the tangent passing through $(a,b)$ is found using the point-slope equation: $$y - b = f'(x)(x-a)$$

But this would be the same equation I would get if I were asked to substitute the point and slope to find the equation of the curve.

Now, I know that the point slope form is for straight lines but I think this should be extendable to any curve because all curves just have variable slopes.

Maybe this is stupid and may not work. Could someone tell me why this is so?

Here's a demonstration of the point slope not yielding back the curve:

$$y=2(x−5)^2 +5 \stackrel{\frac{\mathrm d}{\mathrm dx}\text{ing}}{\implies} y' = 4(x-5)$$ It can clearly be seen that the point $(5,5)$ satisfies the given curve. So, using point-slope: $$ y - 5 = y'(x - 5) \implies y = 4(x−5)^2 +5$$

From this example it is observed that using that the straight line equation does not yield back the curve. My only conclusion is that this is the equation of the tangent passing through $(5,5)$

Please correct where my thinking has gone wrong.

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    $\begingroup$ If we have an equation $$y-b=s(x)\cdot (x-a)$$ for a curve, then $s(x) = \frac{y-b}{x-a}$ is the average slope of the curve between $a$ and $x$. If the slope is constant (the curve is a straight line), the average slope is of course always the pointwise slope. If the curve is not a straight line, then the average slope generally is neither the slope at $a$ nor the slope at $x$. It is, however, the slope at a point between $a$ and $x$: mean value theorem. So for a differentiable function $f$ you have an equation $f(x)-b=f'(\xi(x))\cdot(x-a)$, with a function $\xi$ that usually is complicated. $\endgroup$ – Daniel Fischer Nov 12 '14 at 11:29
  • $\begingroup$ @DanielFischer: Your emphasis on the word "average" leads me to think that I have committed some blunder in equating an instantaneous measure to an average measure. Also, could you give some idea as to where $\xi(x)$ popped out of? $\endgroup$ – Nick Nov 12 '14 at 13:00
  • $\begingroup$ By the mean value theorem, there is a $c$ between $a$ and $x$ such that $\frac{f(x)-f(a)}{x-a} = f'(c)$. For each $x$, choose one of the possible values $c$ and you have your function $\xi$. That function is generally not uniquely determined, but there are such functions. It's not very helpful, however. $\endgroup$ – Daniel Fischer Nov 12 '14 at 13:11
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I'm not sure if you are still looking for clarification on this issue, but here's a go.

For simplicity, let's assume $y=f(x)$ is differentiable for all $x$, and $(x_0, y_0)$ is a point on the curve defined by this equation. (I'll use $x_0$ and $y_0$ instead of $a$ and $b$, for convenience in an explanation below.) The derivative is defined so that $f'(x_0)$ is tangent slope to the curve $y=f(x)$ at the point $(x_0, y_0)$. Then, indeed point-slope form gives us the equation $$ y - y_0 = f'(x_0)(x-x_0). $$

Note what you wrote was not quite right: $y-b = f'(x)(x-a)$ is not true for all $x$, only in general when $x=a$ (so $y=b$ and both sides will be $0$). This was probably the source of your confusion. I believe what you were thinking about is trying to reconstruct the curve $y=f(x)$ from knowing the derivative $f'(x)$ and a specific point $(a,b)$. This is an important topic actually, and is the basic premise behind antiderivatives and differential equations.

Put another way, the equation for the tangent line has 3 components that depend on the specific point: $x_0$, $y_0$ and the slope $f'(x_0)$ at that point, so you can't just replace the $f'(x_0)$ with an arbitrary $f'(x)$ to get the curve, but keep $x_0$ and $y_0$ fixed. (By the way, the curve you got by doing this faulty procedure is not the tangent line either because it's not linear--I don't know that that curve has much meaning.)


Here's a better way to think about trying to reconstruct the curve from the derivative and a specific point. I'll try to explain this as briefly as I can, but feel free to ask for clarification.

Let's work with the simple example of $y=f(x) = x^2$, and pick the starting point $(x_0, y_0) = (0, 0)$. Here $f'(x)= 2x$, so at the starting point the slope is $m_0 = f'(x_0) = 0$. This means the curve is close to the tangent line $y-y_0 = m_0(x-x_0)$, i.e., $y=0$ nearby the point $(0,0)$. That is, we can approximate the curve near $(0,0)$ geometrically by drawing a short horizontal tangent through $(0,0)$.

Now we can do this again. For simplicity, let's say we approximate our curve with the horizontal tangent to $(x_1, y_1) = (1, 0)$ (this is not a point on the curve, but on the tangent line). Now we can use the derivative at 1, $f'(x_1) = f'(1) = 2$, to draw another line segment, say to $(x_2,y_2) = (2,4)$, which is an approximation of the tangent line to $y=f(x)$ at $x_1$.

This is much easier to understand with pictures--take a look at the pictures here for what's going on. I'm just saying we can successively use derivatives (tangent slopes) to approximate the curve. Of course my approximations in my example were very bad, but they get better and better by taking smaller steps (e.g., $x_1 = 0.1$ or $0.00001$, and $x_2 = 2x_1$).

To actually recover the curve $y=f(x)$ exactly, you need to take infinitely small steps, or more precisely, take the limit of doing this process with smaller and smaller steps. This is precisely what the antiderivative is.

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  • $\begingroup$ "trying to reconstruct the curve y = f(x) from knowing the derivative f′(x) and a specific point (a,b)" was exactly what I was attempting. I was an idiot to not see that $y = f'(a)(x-a) + b$. If I had done the math correctly, the result of my example would have been $y = 5$ instead of $y= 4(x-5)^2 + 5$ and this would have been right since I've plugged in the point $(5,5)$. This answer means a lot to me and it has possibly re-established my foundation in straight lines and derivatives. Thank you very much :D $\endgroup$ – Nick Jan 2 '15 at 15:12

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