AboutScotto Expertise Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.).
I also have answered some questions in
Physics (mass, momentum, falling bodies),
Chemistry (charge, reactions, symbols, molecules), and
Biology.
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.
Publications Maybe not a publication, but I have respond to well oveer 3000 questions on the PC.
That's around 2,000 in basic math and 1,000 in advanced math.
Education/Credentials I aquired well over 40 hours of upper division courses. This was well over the number that were required.
I graduated with honors in both my BS and MS degree from Oregon State University.
I was allowed to jump into a few junior level courses my sophomore year.
Awards and Honors I have been nominated as the expert of the month several times.
All of my scores right now are at least a 9.8 average (out of 10).
Past/Present Clients My past clients have been students at OSU, students at the college in South Seattle,
referals from a company, friends and aquantenances, people from my church, and people like you.
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.
