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Learn how derivatives give tangent line equations through slope calculations, visual examples, and curve problems.
---
## Relationship Between Derivatives and Tangent Lines
You probably already know that derivatives can tell us a lot about the properties of a function. One of the coolest uses of derivatives is to find the **slope** or **gradient** of a tangent line to a curve.
Imagine "zooming in" on a point on a curve over and over. Eventually, the curved line will start to look like a straight line, right? Well, that imaginary straight line is what we call the **tangent line**. The gradient of the tangent line at a point $$x=a$$ on the curve $$y=f(x)$$ is exactly equal to the derivative of the function at that point, which is .
Visible text: Imagine "zooming in" on a point on a curve over and over. Eventually, the curved line will start to look like a straight line, right? Well, that imaginary straight line is what we call the **tangent line**. The gradient of the tangent line at a point on the curve is exactly equal to the derivative of the function at that point, which is .
## Determining the Equation of a Tangent Line
To create the equation of a straight line, we need two main things: a point the line passes through and the gradient of the line itself. In this context:
1. **Point of Tangency:** A point $$P(a, b)$$ where the line touches the curve.
2. **Gradient (m):** The slope of the line at that point, which we get from the derivative, .
Visible text: 1. **Point of Tangency:** A point where the line touches the curve.
2. **Gradient (m):** The slope of the line at that point, which we get from the derivative, .
Once we have both, we can plug them directly into the basic formula for a line equation that you're already familiar with:
```math
y - b = m(x - a)
```
> In short, to find the equation of a tangent line, first find its gradient by differentiating the function, then plug the point of tangency and the gradient into the line equation formula.
## Breaking Down a Case
Let's see how it works with an example. Determine the equation of the tangent line to the parabola $$y = 2 - 4x^2$$ at the point $$(1, -2)$$.
Visible text: Let's see how it works with an example. Determine the equation of the tangent line to the parabola at the point .
**Solution:**
**Step** $$1$$: Find the gradient of the tangent line
Visible text: **Step** : Find the gradient of the tangent line
First, we differentiate the function $$f(x) = 2 - 4x^2$$ to get its gradient expression.
Visible text: First, we differentiate the function to get its gradient expression.
```math
f'(x) = -8x
```
Since we want to find the gradient at the point where the $$x$$-coordinate is $$x=1$$, we substitute this value into the derivative.
Visible text: Since we want to find the gradient at the point where the -coordinate is , we substitute this value into the derivative.
```math
m = f'(1) = -8(1) = -8
```
So, the gradient of the tangent line is $$-8$$.
Visible text: So, the gradient of the tangent line is .
**Step** $$2$$: Construct the equation
Visible text: **Step** : Construct the equation
Now we have everything we need:
- Point of tangency $$(a, b) = (1, -2)$$
- Gradient $$m = -8$$
Visible text: - Point of tangency
- Gradient
Plug them into the line equation formula:
Component: MathContainer
Children:
```math
y - b = m(x - a)
```
```math
y - (-2) = -8(x - 1)
```
```math
y + 2 = -8x + 8
```
```math
y = -8x + 6
```
So, the equation of the tangent line is $$y = -8x + 6$$.
Visible text: So, the equation of the tangent line is .
Component: LineEquation
Props:
- title: Visualization of the Tangent Line on the Parabola
- description: This graph shows the parabola curve $$y = 2 - 4x^2$${" "}
and its tangent line $$y = -8x + 6$$ meeting
exactly at the point of tangency $$(1, -2)$$.
Visible text: This graph shows the parabola curve {" "}
and its tangent line meeting
exactly at the point of tangency .
- showZAxis: false
- cameraPosition: [0, 0, 15]
- data: [
{
points: Array.from({ length: 41 }, (_, i) => {
const x = -2 + i * 0.1;
return { x, y: 2 - 4 * x * x, z: 0 };
}),
color: getColor("LIME"),
showPoints: false,
labels: [
{
text: "y = 2 - 4x^2",
at: 5,
offset: [-2, 3, 0],
},
],
},
{
points: [
{ x: 0, y: 6, z: 0 },
{ x: 2, y: -10, z: 0 },
],
color: getColor("YELLOW"),
smooth: false,
showPoints: false,
labels: [
{
text: "y = -8x + 6",
at: 0,
offset: [2.5, -2.5, 0],
},
],
},
{
points: [{ x: 1, y: -2, z: 0 }],
color: getColor("ORANGE"),
showPoints: true,
labels: [
{
text: "(1, -2)",
offset: [1.5, -0.5, 0],
},
],
},
]
## Exercises
1. Find the equation of the tangent line to the curve $$y = x^2 - 6\frac{1}{5}x + 14\frac{1}{2}$$ which is parallel to the line $$x + 2y + 3 = 0$$.
2. Determine the equation of the tangent line to the parabola $$y = 2x^2 - 3x + 5$$ at the point with an ordinate of $$4$$.
3. A curve $$y = 3x - \frac{3}{x^2}$$ intersects the $$x$$-axis at P. Find the equation of the tangent line to the curve at point $$P$$!
4. The curve $$y = (x^2 + 2)^2$$ intersects the $$y$$-axis at point $$A$$. Show that the tangent line to the curve at point $$A$$ is parallel to the $$x$$-axis and is $$4 \text{ units}$$ away from the origin!
5. Determine the coordinates of the point on the curve $$y = 2x^2 - 7x + 1$$, if the tangent line to the curve at that point forms an angle of $$45^\circ$$ with the positive $$x$$-axis. Also, determine the equation of the tangent line to the curve that passes through that point!
Visible text: 1. Find the equation of the tangent line to the curve which is parallel to the line .
2. Determine the equation of the tangent line to the parabola at the point with an ordinate of .
3. A curve intersects the -axis at P. Find the equation of the tangent line to the curve at point !
4. The curve intersects the -axis at point . Show that the tangent line to the curve at point is parallel to the -axis and is away from the origin!
5. Determine the coordinates of the point on the curve , if the tangent line to the curve at that point forms an angle of with the positive -axis. Also, determine the equation of the tangent line to the curve that passes through that point!
### Answer Key
1. **Solution:**
**Step** $$1$$: Determine the gradient.
The tangent line must be parallel to the line $$x + 2y + 3 = 0$$. To find its gradient, let's first convert this line's equation into the slope-intercept form $$y = mx + c$$.
```math
2y = -x - 3
```
```math
y = -\frac{1}{2}x - \frac{3}{2}
```
From this, we know the gradient of the line is $$m = -\frac{1}{2}$$. Since the tangent line is parallel, its gradient is the same.
**Step** $$2$$: Find the point of tangency.
The gradient of the curve at a point is equal to the value of the first derivative at that point. First, let's find the derivative function of $$f(x) = x^2 - \frac{31}{5}x + \frac{29}{2}$$.
Next, we set this derivative equal to the gradient we know ($$m = -1/2$$) to find the $$x$$-coordinate of the point of tangency.
```math
2x - \frac{31}{5} = -\frac{1}{2}
```
```math
2x = \frac{31}{5} - \frac{1}{2} = \frac{62 - 5}{10} = \frac{57}{10}
```
```math
x = \frac{57}{20}
```
After getting the $$x$$-coordinate $$x=\frac{57}{20}$$, we substitute this value back into the *original* curve equation to find its $$y$$-coordinate.
```math
y = (\frac{57}{20})^2 - \frac{31}{5}(\frac{57}{20}) + \frac{29}{2} = \frac{3249}{400} - \frac{1767}{100} + \frac{29}{2} = \frac{3249 - 7068 + 5800}{400} = \frac{1981}{400}
```
So, the point of tangency is $$\left(\frac{57}{20}, \frac{1981}{400}\right)$$.
**Step** $$3$$: Construct the line equation.
With the point $$\left(\frac{57}{20}, \frac{1981}{400}\right)$$ and gradient $$m = -\frac{1}{2}$$, the equation is:
```math
y - \frac{1981}{400} = -\frac{1}{2}\left(x - \frac{57}{20}\right)
```
```math
y = -\frac{1}{2}x + \frac{57}{40} + \frac{1981}{400}
```
```math
y = -\frac{1}{2}x + \frac{570 + 1981}{400}
```
```math
y = -\frac{1}{2}x + \frac{2551}{400}
```
2. **Solution:**
**Step** $$1$$: Find the point(s) of tangency.
Since the ordinate is $$4$$, we set $$y=4$$ in the parabola's equation.
```math
4 = 2x^2 - 3x + 5
```
```math
2x^2 - 3x + 1 = 0
```
```math
(2x-1)(x-1) = 0
```
From factorization, we get two x-values: $$x=1$$ and $$x=\frac{1}{2}$$. This means there are two points of tangency: $$(1, 4)$$ and $$\left(\frac{1}{2}, 4\right)$$. Therefore, there will be two tangent line equations.
**Step** $$2$$: Calculate the gradient and create the equation for each point.
We will process each point of tangency separately. The derivative of the function is .
**Case One: Point $$(1, 4)$$**
The gradient at this point is .
Thus, the equation is:
```math
y - 4 = 1(x - 1) \implies y = x + 3
```
**Case Two: Point $$\left(\frac{1}{2}, 4\right)$$**
The gradient at this point is .
Thus, the equation is:
```math
y - 4 = -1(x - \frac{1}{2}) \implies y = -x + \frac{9}{2}
```
3. **Solution:**
**Step $$1$$: Find point $$P$$.**
The curve intersects the $$x$$-axis when $$y=0$$.
```math
0 = 3x - \frac{3}{x^2}
```
```math
3x = \frac{3}{x^2} \implies 3x^3 = 3 \implies x^3 = 1 \implies x=1
```
So, the intersection point $$P$$ is $$(1, 0)$$.
**Step** $$2$$: Find the gradient at P.
Let's first rewrite the function as $$f(x) = 3x - 3x^{-2}$$ to make it easier to differentiate.
The gradient at $$x=1$$ is .
**Step** $$3$$: Construct the equation.
With point $$(1, 0)$$ and gradient $$9$$, the equation is:
```math
y - 0 = 9(x - 1) \implies y = 9x - 9
```
4. **Solution:**
**Step $$1$$: Find point $$A$$.**
The curve intersects the $$y$$-axis when $$x=0$$.
```math
y = (0^2 + 2)^2 = 4
```
So, the intersection point $$A$$ is $$(0, 4)$$.
**Step $$2$$: Prove the tangent line is parallel to the $$x$$-axis.**
A line parallel to the $$x$$-axis must have a gradient of $$0$$. Let's prove that the derivative of the function at point $$A$$ ($$x=0$$) is zero.
The derivative of $$f(x)=(x^2+2)^2$$ using the chain rule is .
The gradient at $$x=0$$ is .
Since the gradient is zero, **it is proven that the tangent line is parallel to the $$x$$-axis**.
**Step $$3$$: Prove its distance is $$4 \text{ units}$$ from the origin.**
The equation of the tangent line at point $$(0,4)$$ with gradient $$m=0$$ is:
```math
y - 4 = 0(x - 0) \implies y = 4
```
The line $$y=4$$ is a horizontal line. The distance from any point on this line to the $$x$$-axis (the line $$y=0$$) is $$4 \text{ units}$$. Since the origin $$(0,0)$$ lies on the $$x$$-axis, the distance from this tangent line to the origin is also $$4 \text{ units}$$. **Proven.**
5. **Solution:**
**Step** $$1$$: Determine the gradient from the angle.
The relationship between the gradient ($$m$$) and the angle ($$\theta$$) a line makes with the positive $$x$$-axis is given by $$m = \tan(\theta)$$.
```math
m = \tan(45^\circ) = 1
```
So, the gradient of the tangent line we are looking for is $$1$$.
**Step** $$2$$: Find the coordinates of the point of tangency.
The gradient is also the first derivative of the curve $$f(x) = 2x^2 - 7x + 1$$.
We set it equal to the gradient we found:
```math
4x - 7 = 1
```
```math
4x = 8 \implies x = 2
```
Now, find the $$y$$-value by plugging $$x=2$$ into the curve's equation:
```math
y = 2(2)^2 - 7(2) + 1 = 8 - 14 + 1 = -5
```
So, the coordinates of the point of tangency are $$(2, -5)$$.
**Step** $$3$$: Determine the equation of the tangent line.
Using the point $$(2, -5)$$ and gradient $$m=1$$:
```math
y - (-5) = 1(x - 2)
```
```math
y + 5 = x - 2
```
```math
y = x - 7
```
Visible text: 1. **Solution:**
**Step** : Determine the gradient.
The tangent line must be parallel to the line . To find its gradient, let's first convert this line's equation into the slope-intercept form .
From this, we know the gradient of the line is . Since the tangent line is parallel, its gradient is the same.
**Step** : Find the point of tangency.
The gradient of the curve at a point is equal to the value of the first derivative at that point. First, let's find the derivative function of .
Next, we set this derivative equal to the gradient we know () to find the -coordinate of the point of tangency.
After getting the -coordinate , we substitute this value back into the *original* curve equation to find its -coordinate.
So, the point of tangency is .
**Step** : Construct the line equation.
With the point and gradient , the equation is:
2. **Solution:**
**Step** : Find the point(s) of tangency.
Since the ordinate is , we set in the parabola's equation.
From factorization, we get two x-values: and . This means there are two points of tangency: and . Therefore, there will be two tangent line equations.
**Step** : Calculate the gradient and create the equation for each point.
We will process each point of tangency separately. The derivative of the function is .
**Case One: Point **
The gradient at this point is .
Thus, the equation is:
**Case Two: Point **
The gradient at this point is .
Thus, the equation is:
3. **Solution:**
**Step : Find point .**
The curve intersects the -axis when .
So, the intersection point is .
**Step** : Find the gradient at P.
Let's first rewrite the function as to make it easier to differentiate.
The gradient at is .
**Step** : Construct the equation.
With point and gradient , the equation is:
4. **Solution:**
**Step : Find point .**
The curve intersects the -axis when .
So, the intersection point is .
**Step : Prove the tangent line is parallel to the -axis.**
A line parallel to the -axis must have a gradient of . Let's prove that the derivative of the function at point () is zero.
The derivative of using the chain rule is .
The gradient at is .
Since the gradient is zero, **it is proven that the tangent line is parallel to the -axis**.
**Step : Prove its distance is from the origin.**
The equation of the tangent line at point with gradient is:
The line is a horizontal line. The distance from any point on this line to the -axis (the line ) is . Since the origin lies on the -axis, the distance from this tangent line to the origin is also . **Proven.**
5. **Solution:**
**Step** : Determine the gradient from the angle.
The relationship between the gradient () and the angle () a line makes with the positive -axis is given by .
So, the gradient of the tangent line we are looking for is .
**Step** : Find the coordinates of the point of tangency.
The gradient is also the first derivative of the curve .
We set it equal to the gradient we found:
Now, find the -value by plugging into the curve's equation:
So, the coordinates of the point of tangency are .
**Step** : Determine the equation of the tangent line.
Using the point and gradient :