Problems based on tangents to a curve
Problems Based on Tangents to a Curve
Understanding the concept of tangents to a curve is crucial in the field of calculus, particularly when dealing with applications of derivatives. A tangent to a curve at a given point is a straight line that touches the curve at that point without crossing it. This line represents the instantaneous direction of the curve at that point and has the same slope as the curve at that point.
Key Concepts
- Tangent Line: A line that touches a curve at a point without crossing it.
- Point of Tangency: The point at which the tangent touches the curve.
- Slope of the Tangent: The derivative of the curve's equation at the point of tangency.
Finding the Equation of a Tangent Line
To find the equation of a tangent line to a curve at a given point, you need to follow these steps:
- Find the derivative of the curve's equation to determine the slope function.
- Evaluate the derivative at the point of tangency to find the slope of the tangent.
- Use the point-slope form of a line to write the equation of the tangent.
The point-slope form of a line is given by:
[ y - y_1 = m(x - x_1) ]
where ( (x_1, y_1) ) is the point of tangency and ( m ) is the slope of the tangent.
Table: Differences and Important Points
| Aspect | Curve | Tangent Line |
|---|---|---|
| Definition | A set of points that form a continuous path. | A straight line that touches the curve at a single point. |
| Representation | Can be represented by an equation ( y = f(x) ). | Represented by a linear equation ( y = mx + b ). |
| Slope | Varies along the curve. | Constant, equal to the derivative of the curve at the point of tangency. |
| Point of Intersection | N/A | The point of tangency where the curve and the tangent meet. |
| Derivative at a Point | Gives the slope of the tangent at that point. | Not applicable as tangents have a constant slope. |
Examples
Example 1: Find the Tangent to a Parabola
Find the equation of the tangent to the curve ( y = x^2 ) at the point ( (2, 4) ).
Solution:
- Find the derivative of ( y = x^2 ), which is ( \frac{dy}{dx} = 2x ).
- Evaluate the derivative at ( x = 2 ) to find the slope of the tangent: ( m = 2(2) = 4 ).
- Use the point-slope form with ( (x_1, y_1) = (2, 4) ) and ( m = 4 ):
[ y - 4 = 4(x - 2) ]
[ y = 4x - 8 ]
The equation of the tangent is ( y = 4x - 8 ).
Example 2: Tangent Line with a Given Slope
Find the points on the curve ( y = x^3 ) where the tangent line has a slope of 9.
Solution:
- Find the derivative of ( y = x^3 ), which is ( \frac{dy}{dx} = 3x^2 ).
- Set the derivative equal to the given slope and solve for ( x ):
[ 3x^2 = 9 ]
[ x^2 = 3 ]
[ x = \pm\sqrt{3} ]
- Find the corresponding ( y ) values by plugging ( x ) back into the original equation:
[ y = (\pm\sqrt{3})^3 = \pm3\sqrt{3} ]
The points of tangency are ( (\sqrt{3}, 3\sqrt{3}) ) and ( (-\sqrt{3}, -3\sqrt{3}) ).
Practice Problems
- Find the equation of the tangent to the curve ( y = \sqrt{x} ) at ( x = 4 ).
- Determine the point(s) on the curve ( y = \frac{1}{x} ) where the tangent is horizontal.
- For the curve ( y = \sin(x) ), find the equation of the tangent line at ( x = \frac{\pi}{4} ).
By understanding these concepts and practicing problems based on tangents to a curve, students can prepare effectively for exams that include questions on applications of derivatives.