Tangent Line

Tangent: In geometry, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that "just touches" the curve at that point.

key concept 1

The equation for the slope of the tangent line to $f(x)$ is $f^\prime(x)$ .

$f^\prime(x)$ is the derivative of $f(x)$ .

You can always get the slope of the tangent line by using $f^\prime(x)$ .

f(x) is nothing but a function or formula.

key concept 2

And a line equation could be represent as:

the slope-intercept formula: $y = mx + b$ (Where $m$ is the slope of the line, and $b$ is the y-intercept)

The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept.

the point-slope formula: $y - y_1 = m(x - x_1)$ (This formula uses a point on the line, denoted by $(x_1, y_1)$ , and the slope of the line, denoted by $m$ , to calculate the slope-intercept formula for the line)

Point-slope is the general form y-y₁=m(x-x₁) for linear equations. It emphasizes the slope of the line and a point on the line.

Question and Answers

`Area of a triangle` is $18$ . The triangle is enclosed by a `Cartesian coordinate system` and `a tangent`. The equation of the tangent to the curve $y = x^{-\frac{1}{2}}$ is at the point $(a, a^{-\frac{1}{2}})$ . What's the value of $a$ ?

The answer is 64

we should try to understand the question sentence by sentence first.

1.Area of a triangle is :

2. The triangle is enclosed by a Cartesian coordinate system and a tangent.

tangent is a line
the triangle was created by the Cartesian coordinate system and the tangent line

3. The equation of the tangent to the curve is at the point .

The slope of the tangent line is:

\begin{align*} \text{Let's say } f(x) &= x^{-\frac{1}{2}} \\ \\ slope = f^\prime(x) &= -\frac{1}{2}x^{-\frac{1}{2}-1} \\ &= -\frac{1}{2}x^{-\frac{3}{2}} \\ \\ slope = f^\prime(a) &= -\frac{1}{2}a^{-\frac{3}{2}} \end{align*}

Now we already have a point and a slope, then we can generate the equation of the line:

\begin{align*} y - y_1 &= m(x - x_1) \\ \\ y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} )(x - a) \end{align*}

If we could get the x, y in the following picture, we can find an equation of the area of the triangle:

\begin{align*} y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(x - a) \\ \\ \text{when x = 0: } & \\ y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(0 - a) \\ y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(- a) \\ y - a^{-\frac{1}{2}} &= \frac{1}{2}a^{-\frac{3}{2} + 1} \\ y - a^{-\frac{1}{2}} &= \frac{1}{2}a^{-\frac{1}{2}} \\ y &= \frac{1}{2}a^{-\frac{1}{2}} + a^{-\frac{1}{2}} \\ y &= \frac{3}{2}a^{-\frac{1}{2}} \\ \\ \text{when y = 0: } & \\ 0 - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(x - a) \\ - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(x - a) \\ - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) - (-\frac{1}{2}a^{-\frac{3}{2}} \cdot a) \\ - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) + (\frac{1}{2}a^{-\frac{3}{2}} \cdot a) \\ - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) + (\frac{1}{2}a^{-\frac{3}{2} + 1}) \\ - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) + (\frac{1}{2}a^{-\frac{1}{2}}) \\ - a^{-\frac{1}{2}} - (\frac{1}{2}a^{-\frac{1}{2}}) &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) \\ (-1 - \frac{1}{2})a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) \\ (- \frac{3}{2})a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) \\ - \frac{3}{2}a^{-\frac{1}{2}} &= -\frac{1}{2}a^{-\frac{3}{2}} \cdot x \\ \frac{- \frac{3}{2}a^{-\frac{1}{2}}}{-\frac{1}{2}a^{-\frac{3}{2}}} &= x \\ (-\frac{3}{2} \cdot -\frac{2}{1})\frac{a^{-\frac{1}{2}}}{a^{-\frac{3}{2}}} &= x \\ (-\frac{3}{2} \cdot -2)\frac{a^{-\frac{1}{2}}}{a^{-\frac{3}{2}}} &= x \\ 3 \cdot \frac{a^{-\frac{1}{2}}}{a^{-\frac{3}{2}}} &= x \\ 3 \cdot a^{(-\frac{1}{2}) - (-\frac{3}{2})} &= x \\ 3 \cdot a^{(-\frac{1}{2}) + (\frac{3}{2})} &= x \\ 3 \cdot a^1 &= x \\ 3 \cdot a &= x \\ 3a &= x \\ \\ \text{area of triangle} &= \frac{1}{2} \cdot height \cdot width \\ 18 &= \frac{1}{2} \cdot y \cdot x \\ 18 &= \frac{1}{2} \cdot \frac{3}{2}a^{-\frac{1}{2}} \cdot 3a \\ 18 &= (\frac{1}{2} \cdot \frac{3}{2} \cdot 3) \cdot a^{-\frac{1}{2}} \cdot a \\ 18 &= (\frac{9}{4}) \cdot a^{-\frac{1}{2}} \cdot a \\ 18 &= (\frac{9}{4}) \cdot a^{-\frac{1}{2} + 1} \\ 18 &= (\frac{9}{4}) \cdot a^{\frac{1}{2}} \\ 18 \cdot \frac{4}{9} &= a^{\frac{1}{2}} \\ 8 &= a^{\frac{1}{2}} \\ 8 &= \sqrt{a} \\ 8^2 &= (\sqrt{a})^2 \\ 64 &= a \\ a &= 64 \\ \end{align*}

So, finally, we have got the value of a, which is 64.

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Question and Answers

Area of a triangle is

18

. The triangle is enclosed by a Cartesian coordinate system and a tangent. The equation of the tangent to the curve

y = x^{-\frac{1}{2}}

is at the point

(a, a^{-\frac{1}{2}})

. What's the value of

a

The answer is 64

we should try to understand the question sentence by sentence first.

1.Area of a triangle is :

2. The triangle is enclosed by a Cartesian coordinate system and a tangent.

tangent is a line

the triangle was created by the Cartesian coordinate system and the tangent line

3. The equation of the tangent to the curve is at the point .

The slope of the tangent line is:

\begin{align*} \text{Let's say } f(x) &= x^{-\frac{1}{2}} \\ \\ slope = f^\prime(x) &= -\frac{1}{2}x^{-\frac{1}{2}-1} \\ &= -\frac{1}{2}x^{-\frac{3}{2}} \\ \\ slope = f^\prime(a) &= -\frac{1}{2}a^{-\frac{3}{2}} \end{align*}

Now we already have a point and a slope, then we can generate the equation of the line:

\begin{align*} y - y_1 &= m(x - x_1) \\ \\ y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} )(x - a) \end{align*}

If we could get the x, y in the following picture, we can find an equation of the area of the triangle:

\begin{align*} y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(x - a) \\ \\ \text{when x = 0: } & \\ y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(0 - a) \\ y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(- a) \\ y - a^{-\frac{1}{2}} &= \frac{1}{2}a^{-\frac{3}{2} + 1} \\ y - a^{-\frac{1}{2}} &= \frac{1}{2}a^{-\frac{1}{2}} \\ y &= \frac{1}{2}a^{-\frac{1}{2}} + a^{-\frac{1}{2}} \\ y &= \frac{3}{2}a^{-\frac{1}{2}} \\ \\ \text{when y = 0: } & \\ 0 - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(x - a) \\ - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(x - a) \\ - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) - (-\frac{1}{2}a^{-\frac{3}{2}} \cdot a) \\ - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) + (\frac{1}{2}a^{-\frac{3}{2}} \cdot a) \\ - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) + (\frac{1}{2}a^{-\frac{3}{2} + 1}) \\ - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) + (\frac{1}{2}a^{-\frac{1}{2}}) \\ - a^{-\frac{1}{2}} - (\frac{1}{2}a^{-\frac{1}{2}}) &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) \\ (-1 - \frac{1}{2})a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) \\ (- \frac{3}{2})a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) \\ - \frac{3}{2}a^{-\frac{1}{2}} &= -\frac{1}{2}a^{-\frac{3}{2}} \cdot x \\ \frac{- \frac{3}{2}a^{-\frac{1}{2}}}{-\frac{1}{2}a^{-\frac{3}{2}}} &= x \\ (-\frac{3}{2} \cdot -\frac{2}{1})\frac{a^{-\frac{1}{2}}}{a^{-\frac{3}{2}}} &= x \\ (-\frac{3}{2} \cdot -2)\frac{a^{-\frac{1}{2}}}{a^{-\frac{3}{2}}} &= x \\ 3 \cdot \frac{a^{-\frac{1}{2}}}{a^{-\frac{3}{2}}} &= x \\ 3 \cdot a^{(-\frac{1}{2}) - (-\frac{3}{2})} &= x \\ 3 \cdot a^{(-\frac{1}{2}) + (\frac{3}{2})} &= x \\ 3 \cdot a^1 &= x \\ 3 \cdot a &= x \\ 3a &= x \\ \\ \text{area of triangle} &= \frac{1}{2} \cdot height \cdot width \\ 18 &= \frac{1}{2} \cdot y \cdot x \\ 18 &= \frac{1}{2} \cdot \frac{3}{2}a^{-\frac{1}{2}} \cdot 3a \\ 18 &= (\frac{1}{2} \cdot \frac{3}{2} \cdot 3) \cdot a^{-\frac{1}{2}} \cdot a \\ 18 &= (\frac{9}{4}) \cdot a^{-\frac{1}{2}} \cdot a \\ 18 &= (\frac{9}{4}) \cdot a^{-\frac{1}{2} + 1} \\ 18 &= (\frac{9}{4}) \cdot a^{\frac{1}{2}} \\ 18 \cdot \frac{4}{9} &= a^{\frac{1}{2}} \\ 8 &= a^{\frac{1}{2}} \\ 8 &= \sqrt{a} \\ 8^2 &= (\sqrt{a})^2 \\ 64 &= a \\ a &= 64 \\ \end{align*}

So, finally, we have got the value of a, which is 64.

key concept 1

key concept 2

Question and Answers

Area of a triangle is 181818. The triangle is enclosed by a Cartesian coordinate system and a tangent. The equation of the tangent to the curve y=x−12y = x^{-\frac{1}{2}}y=x−21​ is at the point (a,a−12)(a, a^{-\frac{1}{2}})(a,a−21​). What's the value of aaa?

we should try to understand the question sentence by sentence first.

key concept 1

key concept 2

Question and Answers

Area of a triangle is 181818. The triangle is enclosed by a Cartesian coordinate system and a tangent. The equation of the tangent to the curve y=x−12y = x^{-\frac{1}{2}}y=x−21​ is at the point (a,a−12)(a, a^{-\frac{1}{2}})(a,a−21​). What's the value of aaa?

we should try to understand the question sentence by sentence first.

`Area of a triangle` is $18$ . The triangle is enclosed by a `Cartesian coordinate system` and `a tangent`. The equation of the tangent to the curve $y = x^{-\frac{1}{2}}$ is at the point $(a, a^{-\frac{1}{2}})$ . What's the value of $a$ ?

`Area of a triangle` is $18$ . The triangle is enclosed by a `Cartesian coordinate system` and `a tangent`. The equation of the tangent to the curve $y = x^{-\frac{1}{2}}$ is at the point $(a, a^{-\frac{1}{2}})$ . What's the value of $a$ ?