An ant is walking along the curve x^2+xy+y^2=19. If the ant is moving to the right at the rate of 3 centimeters/second, how fast is the ant moving up or down when the ant reaches the point (2,3)? Specify direction.
2x dx/dt + x dy/dt + y dx/dt + 2y dy/dt = 0
(dy/dt)(x + 2y) = -2x dx/dt - y dx/dt
dy/dt(2 + 6) = -4(3) - 3(3)
dy/dt = -21/8
the ant is moving DOWN at 21/8 cm/s
Well, well, well, an ant on a mathematical journey, how marvelous! Let's do some math tricks, shall we?
To find the ant's speed moving up or down, we need to use the concept of partial derivatives. Don't worry, it's not as complicated as it sounds!
But before we start, I have a joke for you:
Why did the math book look sad?
Because it had too many problems!
Now, let's get back to business. The equation of the curve is x^2+xy+y^2=19. To find the ant's speed moving up or down, we need to find dy/dt, which represents the rate of change of y with respect to time.
To do this, we take the derivative of both sides of the equation with respect to t (time), holding the variables x and y constant. The derivative of x^2+xy+y^2 with respect to t will give us the rate of change of the expression with respect to time.
Let's roll our dice in the land of derivatives and see what we get:
2x(dx/dt) + x(dy/dt) + 2y(dy/dt) = 0
Now, all we need to do is to plug in the coordinates of the point where the ant is at (2,3) and the rate at which the ant is moving to the right (3 cm/s).
So, let's take a moment to appreciate the beauty of calculus and make some calculations:
2(2)(3) + 2(3)(dy/dt) + 2(3)(dy/dt) = 0
12 + 12(dy/dt) + 12(dy/dt) = 0
24(dy/dt) = -12
dy/dt = -12/24
dy/dt = -1/2 cm/s
Ta-da! The ant is moving down at a speed of 1/2 cm/s. So, keep your eyes peeled for any falling ants!
To find the rate at which the ant is moving up or down when it reaches the point (2,3), we need to find the derivative of y with respect to time (dy/dt).
Given the equation of the curve x^2 + xy + y^2 = 19, we can implicitly differentiate it with respect to time (t):
2x(dx/dt) + x(dy/dt) + y(dx/dt) + 2y(dy/dt) = 0
We know that dx/dt = 3 cm/s (the ant is moving to the right at the rate of 3 cm/s).
By substituting the values x = 2, y = 3, and dx/dt = 3 into the equation, we can solve for dy/dt:
2(2)(3) + (2)(dy/dt) + (3)(3) + 2(3)(dy/dt) = 0
12 + 2(dy/dt) + 9 + 6(dy/dt) = 0
8(dy/dt) = -21
(dy/dt) = -21/8
The ant is moving down at a rate of 21/8 cm/s when it reaches the point (2,3).
To determine how fast the ant is moving up or down at the point (2,3) on the curve x^2+xy+y^2=19, we can use the concept of implicit differentiation.
Let's differentiate both sides of the equation with respect to time t:
d/dt(x^2+xy+y^2) = d/dt(19)
Using the chain rule, we can rewrite this as:
2x(dx/dt) + x(dy/dt) + 2y(dy/dt) = 0
Now we need to evaluate the values of x, y, dx/dt, and solve for dy/dt.
Given that the ant is moving to the right at a rate of 3 centimeters/second, this means dx/dt = 3 cm/s.
Since we are interested in finding how fast the ant is moving up or down, we are looking for dy/dt at the point (2,3). Therefore, we substitute x = 2 and y = 3 into our equation:
2(2)(3) + 2(3)(dy/dt) + 2(3)(dy/dt) = 0
Simplifying the equation, we have:
12 + 12(dy/dt) + 12(dy/dt) = 0
24(dy/dt) = -12
Now we can solve for dy/dt:
dy/dt = -12/24
dy/dt = -1/2 cm/s
The negative sign indicates that the ant is moving downward at a rate of 1/2 centimeters per second.