if the sides of a cube are measured with an error of 2% use differentials to estimate the relative error in the volume
df= 3x^2 * dx
df= 3x^2 * 1/50
df= 3/50*x^2
is this right?
v = x^3
dv = 3 x^2 dx agrred
dv/v *100 = percent error in volume
dv/v = 3 x^2/x^3 dx
but dx/x * 100 = percent error in side
dv/v = 3 dx/x
so
dv/v * 100 = 3 dx/x * 100
= 3 * 2% = 6 percent
Thank you!! :-)
No, that is not correct. Let's go through the correct steps to estimate the relative error in the volume of a cube when the sides are measured with an error of 2%.
First, let's define the variables:
Let's denote the length of one side of the cube as "x" (without the error) and "dx" as the error in the measurement of the side length, which is 2% of x.
Now, the volume of a cube is given by V = x^3.
To estimate the relative error in the volume, we'll use differentials. The differential of the volume function can be given by:
dV = (dV/dx) * dx
To find (dV/dx), we differentiate the volume function V = x^3 with respect to x:
(dV/dx) = 3x^2
Now, substituting the values into the differential equation, we have:
dV = 3x^2 * dx
Since dx is the error in the measurement of the side length, which is 2% of x, we can substitute it as:
dx = 0.02x
So, the differential equation becomes:
dV = 3x^2 * 0.02x
Simplifying further:
dV = 0.06x^3
To estimate the relative error in the volume, divide both sides of the equation by the actual volume V = x^3:
(dV / V) = (0.06x^3 / x^3)
Canceling out x^3 terms, we get:
(dV / V) = 0.06
Therefore, the relative error in the volume can be estimated as 0.06 or 6%.