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.
A boat is being pulled by rope towards a dock that is 6 feet above where it is tied. The rope is being pulled in at a rate of 2 feet per second. How fast is the boat approaching the dock when there is 10 feet of rope left? At what rate is the angle from the dock changing at that moment?
Answer It is known that 6²+y²=x² where y is the diagonal and x is the base of the right triangle with the dock at the top corner and the boat at the far end on the bottom. The right angle is below the dock and over from the ship.
Taking the derivative, 0 +2y(dy/dt) = 2x(dx/dt).
It is known that x=8, y=10, and dy/dt=-2. The variable dx/dt can be found using the preceding three values.
Take the angle Θ up at the dock between straight down and the cord.
cos(Θ)=y/6. Take the derivative of both sides with Θ and y as functions of t. You get -sin(Θ)dΘ/dt = 6(dy/dt).
Θ is the angle whose sin is 8/10. dy/dt is -2. Θ can be found by moving the values to the right side and then taking arcsin.
