Anonymous

# Calculus maths?

A machine uses a 26cm long moveable rod to move a piston on a vertical shaft. The coordinates of the ends of the rod are (x,0) and (0,y)

http://img61.imageshack.us/my.php?image=64137992nk...

Question A)

Find the distance y(from 0 to p ) in terms of x

Question B) If the bottom end is being pushed horizontally towards the 0 at a constant rate of 1.5cm/s the speed of the piston in cm/s when bottom piston is 24cm above O

Relevance

Step one: Draw a picture

(0,y) pt A

|\

| \

Piston | \ 26 cm rod

| \

| \

--------- (x,0) pt B

As the rod moves the piston up, pt B slides left while pt A rises. We assume the rod is metal and it's length does not change. What does change is the distance from pt B to the shaft, x, and how far out the piston is (how high pt A is), y.

Step 2: Form equation

We know the constant, 26 cm rod, and the variables, x and y. Also constant is the position of the piston, so the 90 degree angle is constant. Since this is true, the triangle is right, and Pythagorean's Theorem relates x,y and 26

x^2 + y^2 = 26^2

From this, we can solve for y for question A

y = SQRT(26^2 - x^2) (toss the neg values of y due to physical reasons)

Step 3: Differentiate the equation

Since we want the speed of the piston, or dy/dt, differentiate the equation with respect to t...

2*x*dx/dt + 2*y*dy/dt = 0 or

x*dx/dt + y*dy/dt = 0

At the point when the bottom of the piston is 24 cm (y = 24), then dx/dt = -1.5 (negative since it is moving left), and x = 10 (use Pythagorean equation 10^2 + 24^2 = 23^2). Plugging in these values gets

(10)*(-1.5) + 24*dy/dt = 0,

dy/dt = 10*1.5/24 = 5/8 or .625 cm/s

• Question A)

26² = x² + y²

y² = 676 - x²

Question B)

2y dy/dt= -2x dx/dt

dy/dt = -x/y * dx/dt

did you want when y = 24?

26² = x² + y²

26² = x² + 24²

x = 10

dy/dt = -x/y * dx/dt

dy/dt = -10/24 * -1.5 (negative because its moving towards 0)

dy/dt = 15/24

dy/dt = 0.625cm/s