Find an equation for the tangent line to the following curve,
[tex]$h(x)=18 x-2 \ln x$[/tex]
Answer (1)
Find the derivative of the function: h ′ ( x ) = 18 − x 2 .
Evaluate the derivative at a point x 0 to find the slope: m = 18 − x 0 2 .
Find the y-coordinate at x 0 : y 0 = 18 x 0 − 2 ln x 0 .
Use the point-slope form to find the tangent line equation: y = ( 18 − x 0 2 ) x + 2 − 2 ln x 0 .
y = ( 18 − x 0 2 ) x + 2 − 2 ln x 0
Explanation
Problem Analysis We are asked to find the equation of the tangent line to the curve h ( x ) = 18 x − 2 ln x . To do this, we need to find the derivative of the function, evaluate it at a specific point (which is not given in the problem, so we'll leave it as a variable x 0 ), and then use the point-slope form of a line to find the equation of the tangent line.
Finding the Derivative First, we find the derivative of h ( x ) with respect to x :
h ′ ( x ) = d x d ( 18 x − 2 ln x ) = 18 − x 2 This derivative gives us the slope of the tangent line at any point x .
Finding the Slope and y-coordinate Let x 0 be the x-coordinate of the point where we want to find the tangent line. The slope of the tangent line at x 0 is: m = h ′ ( x 0 ) = 18 − x 0 2 The y-coordinate of the point on the curve at x 0 is: y 0 = h ( x 0 ) = 18 x 0 − 2 ln x 0
Finding the Tangent Line Equation Now, we use the point-slope form of a line to find the equation of the tangent line: y − y 0 = m ( x − x 0 ) Substitute the expressions for m and y 0 into the point-slope form: y − ( 18 x 0 − 2 ln x 0 ) = ( 18 − x 0 2 ) ( x − x 0 ) Simplify the equation: y = ( 18 − x 0 2 ) x − ( 18 − x 0 2 ) x 0 + 18 x 0 − 2 ln x 0 y = ( 18 − x 0 2 ) x − 18 x 0 + 2 + 18 x 0 − 2 ln x 0 y = ( 18 − x 0 2 ) x + 2 − 2 ln x 0
Final Answer The equation of the tangent line to the curve h ( x ) = 18 x − 2 ln x is: y = ( 18 − x 0 2 ) x + 2 − 2 ln x 0 where x 0 is the x-coordinate of the point of tangency.
Examples
Understanding tangent lines is crucial in physics, especially when analyzing motion. For instance, if h ( x ) represents the position of a particle at time x , the tangent line at a specific time x 0 gives the instantaneous velocity of the particle at that moment. This concept is fundamental in understanding how objects move and change their velocity over time, providing a direct application of calculus in real-world scenarios.
