Find an equation of the tangent line to the curve at the given point y = x, (81, 9) involves analyzing a radical function and applying derivative concepts. This walkthrough shows how to determine the precise linear approximation at the specified coordinate on the curve.
Working with expressions that involve roots and their inverses requires careful algebraic manipulation. The process connects limits, slope calculations, and point-slope form to produce a reliable mathematical model for instantaneous change.
| Function | Given Point | Goal | Method |
|---|---|---|---|
| y = x | (81, 9) | Tangent line equation | Derivative and point-slope form |
| Rewrite as y = x^(1/2) | x = 81 | Confirm domain | Power rule preparation |
| Curve behavior | y = 9 at x = 81 | Verify point lies on curve | Substitution check |
| Derivative expression | f'(x) = 1/(2√x) | Slope at given x | Instantaneous rate |
Derivative Evaluation at the Given Coordinate
To find the slope of the tangent line, compute the derivative of y = x using exponent and chain considerations. Substituting x = 81 into f'(x) yields a precise numerical rate of change.
Simplifying the Derivative Expression
Rewrite the radical as a fractional exponent before differentiating. Apply the power rule and simplify the resulting algebraic fraction to a clean form suitable for evaluation.
Computing the Numerical Slope
Plug x = 81 into the derivative to obtain m = 1/18. This fraction represents the steepness of the tangent line at the designated point on the curve.
Point-Slope Construction and Simplification
With slope m = 1/18 and the coordinate (81, 9), apply the point-slope template to generate the initial linear relationship. Rearrange terms systematically to isolate y and express the result in slope-intercept format.
Arranging Terms for Clarity
Distribute the fraction and combine constants to eliminate denominators. Ensure all variables remain on one side and rational coefficients are expressed in lowest terms.
Verification and Graphical Consistency
Confirm that the derived line passes through the original point and that its slope aligns with the curve's behavior near x = 81. Visual checks help catch algebraic slips and validate the model.
Testing the Line Equation
Substitute x = 81 into the tangent equation to verify that y returns exactly 9. Evaluate small deviations around the point to observe how closely the line approximates the curve.
Summary of Tangent Line Process
- Confirm the point satisfies the original function y = x.
- Differentiate to find f'(x) = 1/(2√x).
- Evaluate the derivative at x = 81 to obtain slope m = 1/18.
- Apply point-slope form using (81, 9) and the computed slope.
- Simplify to slope-intercept or standard linear form.
- Verify the line passes through the point and matches local curve behavior.
FAQ
Reader questions
How do I verify that the point lies on the curve y = x?
Substitute x = 81 into the function to obtain y = 9, confirming that (81, 9) satisfies the equation and lies on the curve.
What is the derivative of y = x at any x > 0?
The derivative is f'(x) = 1/(2√x), which gives the instantaneous slope of the curve at any positive x value.
How do I calculate the slope at x = 81 specifically?
Evaluate f'(81) = 1/(2√81) = 1/18, producing the exact slope required for the tangent line.
Can the final equation be written without fractions?
Yes, by clearing denominators during rearrangement, the equation can be expressed as x - 18y + 81 = 0 while preserving accuracy.