Math Problem Statement
Solution
To solve the problem in Example 5, we aim to find the equation of a tangent line to the curve that passes through the origin .
Step-by-step solution:
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Determine the slope of the tangent line: The derivative of is . This gives the slope of the tangent line at any point on the curve.
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Write the equation of the tangent line: The general equation of a line is: Here, , where is the -coordinate of the point of tangency, since the slope of the curve at is . The point of tangency on the curve is .
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Condition for the tangent line passing through the origin: If the line passes through the origin , substituting and into the line equation yields: Therefore, the line equation simplifies to:
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Point of tangency constraint: The tangent line must also pass through the point of tangency . Substituting this point into the line equation: Simplifying: Since , it follows that:
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Equation of the tangent line: The point of tangency is , and the slope is . The tangent line is:
Final Answer:
The equation of the tangent line is:
Would you like further clarification or additional details? Here are related questions to expand upon:
- What is the significance of finding tangents to a curve in real-world applications?
- Can a tangent line to pass through any other fixed point?
- How does the derivative help in determining tangency?
- Why is unique in terms of its derivative and tangent properties?
- How would the solution change if the curve was instead?
Tip: The exponential function is special because its derivative is the same as the function itself, making its slope at any point straightforward to compute!
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Math Problem Analysis
Mathematical Concepts
Calculus
Exponential Functions
Tangent Lines
Formulas
y = mx + c (Equation of a straight line)
dy/dx = e^x (Derivative of y = e^x)
Theorems
Derivative Rule for Exponential Functions
Suitable Grade Level
Grades 11-12