What Does "Tangent to a Point on a Curve" Mean?
Basically, a tangent line through a given point on a curve is the boundary of a half-plane which contains a segment of the curve to either side of the point, and for which the point itself is on the boundary.
I will give you an example:
Say the curve $y = 5 - x^2$ at the point $(1,4)$ on the curve and you want to find the slope of line tangent to a curve at that given point. You would basically utilize the derivative, $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$ to find the slope of the tangent line to the curve at the point $(1, 4)$, where $$y = f(x) = 5 - x^2$$ So, $$\begin{align} f'(x) =\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} & = \lim_{h \to 0} \frac{\Big(5 - (x + h)^2\Big) - (5 - x^2)}{h} \\ \\ & = \lim_{h\to 0}\frac{5 - (x^2 + 2hx + h^2) - 5 + x^2}{h} \\ \\ & = \lim_{h\to 0}\frac {-2hx - h^2}{h} \\ \\ & = \lim_{h\to 0}\frac{-h(2x + h)}{h} \\ \\ & = \lim_{h\to 0} -(2x + h)\\ \\ f'(x) & = -2x \end{align}$$
Now, at $(1, 4), \;x = 1$, so the slope at that point is given by $f'(1) = -2(1) = -2$.
Here is a depiction of how Wolfram Alpha shows it: (Image compliments of WolframAlpha).
If you want to go further and find the actual equation of the line then, you can find the equation line tangent to $y = 5-x^2$ at the point $(1,4)$ knowing the slope of the line is $-2$, and the fact that $(1, 4)$ lies on that line. We know the equation of a line can be written in the form $(y - y_0) = m(x - x_0)$, where $m$ represents the slope of the line, and $(x_0, y_0)$ represents a point on the line.
In our case, we've found slope to be given by $-2$, and $(x_0, y_0) = (1, 4)$. So, $$y - 4 = -2(x - 1) \iff y - 4 = -2x + 2 \iff y = 6 - 2x$$
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Inquisitive
Updated on September 23, 2022Comments
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Inquisitive about 1 year
Is has been stated in an earlier answer to another question that a "point" is an undefined concept in mathematics.
If this is true, how can we define "tangent to a point on a curve"?
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André Nicolas almost 9 yearsYou may mean tangent to a curve at a point.
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turkeyhundt almost 9 yearsI think this is a case where mathematical terms and plain language(english) might be blurring together. Just because a point is "undefined" doesn't mean it doesn't exist.
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bjd2385 almost 9 yearsPoints certainly exist, but I believe this is really just a mixup. They likely mean the tangent line to a curve, which indicates that it is tangent to one point that is a member of the set of points making up the curve.
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Inquisitive almost 9 yearsIs a curve made up of a string of extremely tiny circles?
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Inquisitive almost 9 yearsI received criticism for an answer I provided elsewhere. I began to question whether or not I understood what a simple point was. As it turns out, even heavy hitters here are confused by the concept. I was hoping for something that would clarify the concept in my mind.
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MPW almost 9 years@IsaacWannabeeNewton: Yes--those extremely tiny circles are circles of radius zero, also known as "points". :/
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Inquisitive almost 9 yearsYagna, is it possible to have two tangent lines at that one point with each of the tangents having a different slope?
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Admin almost 9 years@IsaacWannabeeNewton No it is not possible. However the slope varies with the slightest change on the graph itself. So, if the point is say, (2,1), the slope would be completely different. The slope is basically dependent on the derivative of the function.
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Admin almost 9 years@IsaacWannabeeNewton Well not completely: it would be -4. And this is easily verifiable by the graph. As you move along the curve, the slope changes.
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Admin almost 9 years@IsaacWannabeeNewton No Problem.
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Pete L. Clark almost 9 years"Basically, a tangent line through a given point on a curve is the boundary of a half-plane which contains a segment of the curve to either side of the point, and for which the point itself is on the boundary." This is not a bad first intuition about tangent lines. Subject to suitable differentiability and convexity conditions, it becomes a characteristic property. But it is not suitable as a definition in general: consider even that the tangent line to $f(x) = x^3$ at $x = 0$ does not satisfy your "boundary plane" property.