Answer to Question #269326 in Calculus for Angel Nodado

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.

Answer 1

We have a cube whose edge is 15 cm.

The error in calculating the edge is 0.01 cm

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

 $V=x^3$

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

$A=x^2$

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

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

Part(a)

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

Now, 

$V=x ^3$

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

$\frac{dV}{dx}=3x^2\\⇒dV=3x^2 dx$

Now, x=15 cm and dx=±0.01 cm, therefore, 

$dV=3x^2 dx\\=3(15)^2×(±0.01)\\=675×(±0.01)\\=±6.75$

Hence, the error in computing the volume is $±6.75 cm^3$

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

Now,$A=x ^2$

Differentiating with respect to x on both sides:

$\frac{dA}{dx}=2x\\⇒dA=2x dx$

Now, x=15 cm and dx=±0.01 cm therefore, 

$dA=2x dx\\=2(15)×(±0.01)\\=30×(±0.01)\\=±0.3$

Hence, the error in computing the area of one of the faces is $±0.3 cm^2$