Calculus 2 Help

[kimsworld]

Hello all! My teacher gave this problem about a week ago and I have had no idea how to solve even the first part. I went to the tutors at my school who, most unfortunately, couldn't figure it out either. Please Help!


Problem: Given C is the graph of the equation
2radical3 * sinpi(x)/3 =y^5+5y-3


(1) Prove that as a set
C= {(x,y) Exists at all Real Numbers Squared | 2radical3 * sinpi(x)/3 =y^5+5y-3
is the graph of a function differentiable on all real numbers using the Inverse Function Theorem and Chain Rule.


(2) Prove that C contains the point (1,1)

(3) Obtain the slope and a cartesian equation of the line l tangent to C at the point (1,1). Prove your answer.

(4) A point (x(t), y(t)) moves all along C as t increases in all real numbers with constant horizontal rate with respect to t of 30 units per second
(For all t that exists as real numbers : x'(t) = 30/sec).
What is the vertical rate w.r.t. t when x(t) =1 ([y']x=1)?


*Note: The equation given for C cannot be solved for y using just the standard arithmetic operations, including radicals, but can be solved using inverse function notation.

 

[HallsofIvy]

Hello all! My teacher gave this problem about a week ago and I have had no idea how to solve even the first part. I went to the tutors at my school who, most unfortunately, couldn't figure it out either. Please Help!


Problem: Given C is the graph of the equation
2radical3 * sinpi(x)/3 =y^5+5y-3

Start by writing the equation correctly! The is no such function as "sinpi". Did you mean "sin(pi x)/3" or "sin(pi x/3)"?

(1) Prove that as a set
C= {(x,y) Exists at all Real Numbers Squared | 2radical3 * sinpi(x)/3 =y^5+5y-3
is the graph of a function differentiable on all real numbers using the Inverse Function Theorem and Chain Rule.

The simplest way to show that a function is differentiable is find its derivative! The "inverse function theorem" just says that you can think of this equation as defining y as a function of x. And the chain rule says that the derivative, with respect to x, of y^5+ 5y- 3 is the derivative with respect to y, times the derivative of y with respect to x, y'. Differentiate both sides of the equation with respect to x and then solve for y'. (Does the problem really say "at all Real Numbers squared"?)

(2) Prove that C contains the point (1,1)

If you understand what this means this is just arithmetic. Set x and y equal to 1 in the equation. Is the result true?

(3) Obtain the slope and a cartesian equation of the line l tangent to C at the point (1,1). Prove your answer.

The slope is just y' at (1, 1). The equation of the tangent line is y= m(x- 1)+ 1 where "m" is the slope.

(4) A point (x(t), y(t)) moves all along C as t increases in all real numbers with constant horizontal rate with respect to t of 30 units per second
(For all t that exists as real numbers : x'(t) = 30/sec).
What is the vertical rate w.r.t. t when x(t) =1 ([y']x=1)?

dy/dt= (dy/dx)(dx/dt)= (dy/dx)(30).

*Note: The equation given for C cannot be solved for y using just the standard arithmetic operations, including radicals, but can be solved using inverse function notation.