AboutScotto Expertise Any kind of mathematics (algebra, geometry, trigonometry, matrices, calculus, linear approximation, linear regression, linear programming, numerical analysis, probability, statistics, etc.). I also have answered some questions in Physics, Chemistry, and Biology. I would like to volunteer in all areas of Mathematics, not just calculus, and the other three courses that were mentioned.
Experience Experience in the area: I have tutored students in all areas of mathematics for over 20 years.
Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors.
Awards and Honors: I have passed Actuarial tests 100, 110, and 135.
Expert: Scotto Date: 6/27/2008 Subject: finding the equation of the tangent line
Question My problem is:
The point P(5,2) lies on the curve y=sqrt(x-1). Find an equation of the tangent line to the curve at P(5,2).
I've been taught to solve this by finding the slope of the secant line, however, I'm really unsure of how to go about doing this.
So far, I've graphed the equation and found a second point Q(10,3) and the slope of the line between those to points. So, mPQ = 5. I'm really not sure where to go from here though or even if I'm going in the right direction.
Thanks for your help!
Answer I'm not sure if that's the right direction or not. Here is what I do know.
To find a tangent line, we have a point and we know that it is parallel to the curve at (5,2). Note that x^0.5 is the same as x^(1/2). The slope of the line is the dervative at this point. Since the function is (x-1)^0.5, the derivative is 0.5(x-1)^-0.5 = 1/(2(x-1)^0.5). Put in the x value to find the slope and use (5,2) as (x0,y0). That all fits into the point-slope form of a line.
