A window has the shape of a semicircle. The base of the window is measured as having diameter 64 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum error possible in computing the area of the window. a. 1.3π cm^2
b. 2.4 πcm^2
c. 2.6 πcm^2
d. 3.2 πcm^2
e. 1.6 πcm^2
f. 1.2 πcm^2

e. 1.6π cm²

Step-by-step explanation:

Since the window is in a semi-circular shape, its area A = πD²/4 ÷ 2 = πD²/8 where D = diameter of window = 64 cm

Now, the error in the area dA = dA/dD × dD where dD = error in the diameter = 0.1 cm and dA/dD = derivative of A with respect to D.

So, dA/dD = d(πD²/8)/dD = 2 × πD/8 = πD/4

So, the differential dA = dA/dD × dD

dA =  πD/4 × dD

Substituting D = 64 cm and dD = 0.1 cm into the equation, we have

dA =  πD/4 × dD

dA =  π × 64 cm/4 × 0.1 cm

dA =  π × 16 cm × 0.1 cm

dA =  π × 1.6 cm²

dA = 1.6π cm²

So, the maximum error in computing the area of the window is 1.6π cm²

slope intercept form is y = mx + b