Question

The measurement of an edge of a cube is found to be 15 cm with a
possible error of 0.01 cm. Find the approximate error in computing a.)
the volume. b.) the area of one of the faces.

Accepted Answer

We have a cube whose edge is 15 cm.

The error in calculating the edge is 0.01 cm

0.01 cm

Now, this error can be in addition or subtraction, therefore, the
error is ±0.01 cm

±0.01 cm

We need to find the error in computing the volume and area of one of
the faces.

Now, the volume of a cube with side x cm is:

V=x

3

V=x3

And the area of one of the sides is equal to the area of the square
which is :

A=x

2

A=x2

Step 2

Now, we have the side of the cube is x=15 cm

x=15 cm

The error in computing the edge is dx=±0.01 cm

dx=±0.01 cm

Part(a)

We need to find the error in computing the volume, that is, dV

dV

Now,

V=x

3

V=x3

Differentiating it with respect to x on both the sides we get:

dV

dx

=3x

2

⇒dV=3x

2

dx

dVdx=3x2⇒dV=3x2 dx

Now, x=15 cm

x=15 cm and dx=±0.01 cm

dx=±0.01 cm, therefore,

dV

=

3x

2

dx

3(15)

2

×(±0.01)

675×(±0.01)

±6.75

dV=3x2 dx=3152×±0.01=675×±0.01=±6.75

Hence, the error in computing the volume is ±6.75 cm

3

±6.75 cm3

Answer: The error in computing the volume is ±6.75 cm

3

±6.75 cm3

Step 3

Part(b)

Now, we need to find the error in computing the area of one of the
faces. That means we need to find dA

dA

Now, A=x

2

A=x2

Differentiating with respect to x on both sides:

dA

dx

=2x

⇒dA=2x dx

dAdx=2x⇒dA=2x dx

Now, x=15 cm

x=15 cm and dx=±0.01 cm

dx=±0.01 cm, therefore,

dA

=

2x dx

2(15)×(±0.01)

30×(±0.01)

±0.3

dA=2x dx=215×±0.01=30×±0.01=±0.3

Hence, the error in computing the area of one of the faces
is ±0.3 cm

2

±0.3 cm2

Answer: The error in computing the area of one of the faces
is ±0.3 cm

2

±0.3 cm2

Question #269131

Expert's answer

Answer in progress...