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 .
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:
Point of Tangency: A point P(a,b) where the line touches the curve.
Gradient (m): The slope of the line at that point, which we get from the derivative, m=f′(a).
Once we have both, we can plug them directly into the basic formula for a line equation that you're already familiar with:
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.
Find the equation of the tangent line to the curve y=x2−651x+1421 which is parallel to the line x+2y+3=0.
Determine the equation of the tangent line to the parabola y=2x2−3x+5 at the point with an ordinate of 4.
A curve y=3x−x23 intersects the x-axis at P. Find the equation of the tangent line to the curve at point !
The curve y=(x2+2)2 intersects the y-axis at point A. Show that the tangent line to the curve at point is parallel to the -axis and is away from the origin!
Determine the coordinates of the point on the curve y=2x2−7x+1, if the tangent line to the curve at that point forms an angle of 45∘ with the positive x-axis. Also, determine the equation of the tangent line to the curve that passes through that point!
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.
2y=−x−3
y=−21x−23
From this, we know the gradient of the line is m=−21. Since the tangent line is parallel, its gradient is the same.
Step2: 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)=x2−531x+229.
f′(x)=2x−531
Next, we set this derivative equal to the gradient we know (m=−1/2) to find the x-coordinate of the point of tangency.
2x−531=−21
After getting the x-coordinate x=2057, we substitute this value back into the original curve equation to find its y-coordinate.
From factorization, we get two x-values: x=1 and x=21. This means there are two points of tangency: (1,4) and (21,4). Therefore, there will be two tangent line equations.
Step2: Calculate the gradient and create the equation for each point.
We will process each point of tangency separately. The derivative of the function is f′(x)=4x−3.
Case One: Point (1,4)
The gradient at this point is m=f′(1)=4(1)−3=1.
Thus, the equation is:
y−4=1(x−1)⟹y=x+3
Case Two: Point (21,4)
The gradient at this point is m=f′(21)=4(21)−3=−1.
Thus, the equation is:
y−4=−1(x−21)⟹y=−x+29
0=3x−x23
3x=x23⟹3x3=3⟹x3=1⟹x=1
So, the intersection point P is (1,0).
Step2: Find the gradient at P.
Let's first rewrite the function as f(x)=3x−3x−2 to make it easier to differentiate.
f′(x)=3−(−2)(3)x−3=3+x36
The gradient at x=1 is m=f′(1)=3+136=9.
Step3: Construct the equation.
With point (1,0) and gradient 9, the equation is:
y−0=9(x−1)⟹y=9x−9
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)=(x2+2)2 using the chain rule is f′(x)=2(x2+2)(2x)=4x(x2+2).
The gradient at x=0 is m=f′(0)=4(0)(02+2)=0.
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 units from the origin.
The equation of the tangent line at point (0,4) with gradient m=0 is:
y−4=0(x−0)⟹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 units. Since the origin (0,0) lies on the x-axis, the distance from this tangent line to the origin is also 4 units. Proven.
m=tan(45∘)=1
So, the gradient of the tangent line we are looking for is 1.
Step2: Find the coordinates of the point of tangency.
The gradient is also the first derivative of the curve f(x)=2x2−7x+1.
f′(x)=4x−7
We set it equal to the gradient we found:
4x−7=1
4x=8⟹x=2
Now, find the y-value by plugging x=2 into the curve's equation:
y=2(2)2−7(2)+1=8−14+1=−5
So, the coordinates of the point of tangency are (2,−5).
Step3: Determine the equation of the tangent line.
