The speed s of an object dropped from rest varies with the distance y it has fallen according to the equation s=2gy−−−√ , where g is the acceleration due to gravity. What is the instantaneous rate of change of the speed with respect to distance? (1 point)
To find the instantaneous rate of change of the speed with respect to distance, we need to find the derivative of the speed equation with respect to distance.
Using the power rule and chain rule, we can find:
ds/dy = d/dy (2gy^0.5)
= 2g * 0.5 * y^(-0.5)
= g * y^(-0.5)
Therefore, the instantaneous rate of change of the speed with respect to distance is g * y^(-0.5).
To find the instantaneous rate of change of the speed with respect to distance, we need to differentiate the equation s=2gy−−−√ with respect to y.
Let's differentiate the equation using the chain rule:
s = 2gy^(1/2)
To differentiate s with respect to y, we differentiate the expression inside the square root and then multiply by the derivative of the square root function:
s' = 2g * (1/2) * y^(-1/2)
Simplifying further, we get:
s' = g * y^(-1/2)
Therefore, the instantaneous rate of change of the speed with respect to distance is g multiplied by y to the power of -1/2.
To find the instantaneous rate of change of the speed with respect to distance, we need to take the derivative of the given equation, s = 2√(gy).
First, let's rewrite the equation as s = (2g * y)^1/2.
Now, we can differentiate both sides of the equation with respect to y.
The derivative d/dy of (2g * y)^1/2 is found using the chain rule:
d/dy [(2g * y)^1/2] = (1/2) * (2g * y)^(-1/2) * d/dy (2g * y)
Using the power rule, the derivative of (2g * y) is:
d/dy (2g * y) = 2g
Substituting this back into the equation, we have:
d/dy [(2g * y)^1/2] = (1/2) * (2g * y)^(-1/2) * 2g
Simplifying further:
d/dy [(2g * y)^1/2] = g / (2√(gy))
Therefore, the instantaneous rate of change of the speed with respect to distance is g / (2√(gy)).