Find the equation of the tangent to the curve $y=2 \ln x$ at the point on the curve with $x$-coordinate $x=1$.
Answer (1)
Find the y -coordinate at x = 1 : y = 2 ln ( 1 ) = 0 .
Calculate the derivative: d x d y = x 2 .
Evaluate the derivative at x = 1 to find the slope: m = 1 2 = 2 .
Use the point-slope form to find the tangent line equation: y − 0 = 2 ( x − 1 ) , which simplifies to y = 2 x − 2 .
Explanation
Problem Analysis We are given the curve y = 2 ln x and asked to find the equation of the tangent line at the point where x = 1 .
Finding the y-coordinate First, we need to find the y -coordinate of the point on the curve where x = 1 . Substituting x = 1 into the equation y = 2 ln x , we get y = 2 ln ( 1 ) = 2 ⋅ 0 = 0. So the point of tangency is ( 1 , 0 ) .
Finding the derivative Next, we need to find the derivative of the function y = 2 ln x with respect to x . Using the rule that the derivative of ln x is x 1 , we have d x d y = 2 ⋅ x 1 = x 2 .
Finding the slope Now, we evaluate the derivative at x = 1 to find the slope of the tangent line at the point ( 1 , 0 ) .
m = d x d y x = 1 = 1 2 = 2. So the slope of the tangent line is 2 .
Finding the equation of the tangent line Finally, we use the point-slope form of a line, 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 line, to find the equation of the tangent line. In this case, ( x 1 , y 1 ) = ( 1 , 0 ) and m = 2 , so the equation of the tangent line is y − 0 = 2 ( x − 1 ) y = 2 x − 2.
Final Answer Therefore, the equation of the tangent line to the curve y = 2 ln x at the point where x = 1 is y = 2 x − 2 .
Examples
Imagine you're tracking the growth of a plant, where the height y is modeled by y = 2 ln x , with x representing time. Finding the tangent line at a specific time x = 1 helps you estimate the plant's immediate growth rate at that moment. This is useful for predicting future growth trends and making informed decisions about plant care, such as adjusting watering or sunlight exposure to optimize growth.
