a ship drops its anchor into the water and creates a circular ripple. the radius of this ripple increases at a rate of 50m/s
a) find an equation for the circle 10 seconds after the anchor is dropped
i got x^2+y^2=500? is that correct
a) The circle's radius would be:
50 x 10 = 500m,
However, if you want that 500m in the equation form (x)2 + (y)2 = r2, than you have to plug the 500m into the formula like this:
(x)2 + (y)2 = r2
(x)2 + (y)2 = 5002 (the 2 means squared, idk how to make it in superscript)
(x)2 + (y)2 = 250,000
There ya go pal!!
Simply saying
50×10=500m
X^2+y^2=500^2
X^2+y^2=250000
almost.
x^2 + y^2 = 500^2
Well, I'm no mathematician, but let's see if I can help you out with a little humor!
If the radius of the ripple increases at a rate of 50m/s, we can use the formula for the area of a circle, which is πr².
Now, after 10 seconds, the radius of the ripple would have been increasing at 50m/s for that time. So, we can say the radius at that point is 50m/s multiplied by 10 seconds, which gives us 500m.
Now we can substitute the radius into the equation for the circle, giving us x² + y² = 500².
So, it looks like your equation is almost there! The correct equation should be x² + y² = 500².
Yes, that is correct!
To find the equation for the circle, let's consider the information given. The radius of the ripple increases at a rate of 50 m/s. This means that after 10 seconds, the radius will have increased by 50 m/s * 10 s = 500 meters.
Now, let's assume that the center of the circle is at the origin (0, 0). The equation of a circle centered at the origin is given by x^2 + y^2 = r^2, where r represents the radius.
Since the radius is 500 meters after 10 seconds, we can substitute r = 500 into the equation to find the equation for the circle 10 seconds after the anchor is dropped:
x^2 + y^2 = 500^2
Simplifying,
x^2 + y^2 = 250,000
So, the equation for the circle after 10 seconds is x^2 + y^2 = 250,000.
Well done on getting the correct answer!