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About Scotto
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.

You are here: Experts > Teens > Homework/Study Tips > Calculus > calculus

Calculus - calculus

Expert: Scotto - 11/4/2009

Question
the question ask, to find the point on the graph of the curve given by y^2 = 2x that is closest to point (1,4) stuck on this one... does this have to do with parametric equations and taking the derivative of y^2 = 2x to find slope... a little stuck

Answer
My answer was modified slightly, but it was correct.
If you had any trouble comprehending my 1st answer, here it is revised.
I am positive my 2nd answer looks right.

If we are given y = y0, then x0 = (y0)²/2.

The distance is given by s² = (x-1)² + (y-4)².

The derivative with x as a function of y is 0 = 2(x-1)(dx/dy) + 2(y-4).

Dividide both sides by 2, giving 0 = (x-1)(dx/dy) + y-4.

Add 4-y to both sides, getting 4-y = (x-1)(dx/dy).

Since (x0,y0) is on the line with that slope, we can put in what y0 for y and x0 for x.
Note that x0 was given at the end of the first paragraph. This gives 4 - y0 = ((y0)²/2 - 1)(dx/dy).

Multiply both sides by 2 and get 8 - 2y0 = ((y0)² - 2)(dx/dy).
Divide both sides by ((y0)² - 2). This gives (8-2•y0)/((y0)²-2) = dx/dy.

To minimize it, we set the derivative = to 0 and not that the only thing that matters is the numerator, for the denominator can never make a fraction 0. Thus, we have 8 - 2•y0 = 0, or 8 = 2•y0, or 4 = y0.

Since y0² = 2•x0, then 4² = 16 = 2•x0, so 8 = x0.

Thus, the point on the curve that is closest is (8,4).

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