can you please help me with this? The area of a triangle with sides of length a,

Question

can you please help me with this?

The area of a triangle with sides of length a, b, and c is given by a formula from antiquity called Heron's formula, shown below, where s = (a + b + c) / 2 is the semi perimeter of the triangle. A = s (s - a) (s - b) (s - c) Complete parts a. through c. below. Find the partial derivatives Aa, Ab, and Ac. Find the partial derivative Aa. Choose the correct answer below. Aa = a3 + a2b + ac2/2 s (s - a) (s - b) (s - c) Aa = - a3 + a2b + ac2 Aa = - a3 + ab2 + ac2/8 s (s - a) (s - b) (s - c) Aa = - a3 + a2b + ac2/s (s - a) (s - b) (s - c) Find the partial derivative Ab. Choose the correct answer below. Ab = - b3 + a2b + bc2 Ab = - b3 + ab2 + bc2/2 s (s - a) (s - b) (s - c) Ab = - b3 + ab2 + bc2/ s (s - a) (s - b) (s - c) Ab = - b3 + a2b + bc2/8 s(s - a)(s- b)(s - c) Find the partial derivative Ac, Choose the correct answer below. - c3 + ac2 + b2c/ s (s - a) (s - b) (s - c) Ac = - c3 + a2c + b2c Ac = - c3 + a2c + b2c/8 s (s - a) (s - b) (s - c) Ac = c3 + ac2 + b2c/2 s (s - a) (s - b) (s - c) A triangle has sides of length a = 2, b = 4, c = 5. Using the partial derivatives found in part a, estimate the change in the area when a increases by 0.03, b decreases by 0.08, and c increases by 0.6. The change in area is approximately . (Round to two decimal places as needed.) For an equilateral triangle with a = b = c, estimate the percent change in the area when all sides increase in length by p%. The percent change in area is approximately %. (Simplify your answer.)

Explanation / Answer

F = s(s-a)(s-b)(s-c)

dF/da = a(b^2 + c^2 - a^2)/4

dF/db = b(a^2 + c^2 - b^2)/4

dF/dc = c(a^2 + b^2 - c^2)/4

Therefore:

1)

Aa =  a(b^2 + c^2 - a^2)/4 * 1/2sqrt(s(s-a)(s-b)(s-c)) -> answer C.

2)

Ab = b(a^2 + c^2 - b^2)/4 * 1/2sqrt(s(s-a)(s-b)(s-c)) -> answer D.

3)

Ac = c(b^2 + a^2 - c^2)/4 * 1/2sqrt(s(s-a)(s-b)(s-c)) -> answer C.

b)

a=2,b=4,c=5 -> s = 11/2 -> F = 11/2 * 1/2 * 7/2 * 3/2 -> F = 231/16

Aa =  2(16 + 25 - 4)/4 * 1/2 * 4/sqrt(231) = 2.43

Ab = 4(4+25-16)/4 * 1/2 * 4/sqrt(231) = 1.71

Ac = 5(4+16-25)/4 * 1/2 * 4/sqrt(231) = -0.82

So the change in area would be:

2.43*0.03 + 1.71*(-0.08)-0.82*0.6 = -0.556

c)

Aa = Ab = Ac = a^3/4 * 1/2 * 4/sqrt(3)a^2 = a/2sqrt(3)

-> percentage of area increase is:

(a/2sqrt(3) * p * a * 3)/(a^2 * sqrt(3) / 4) =

(p/2)/(1/4) =

2p

(2p)% increase in area

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