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Yuliya22 [10]
3 years ago
10

Show that A(t)=300−250e0.2−0.02t satisfies the differential equation ⅆAⅆt=6−0.02A with initial condition A(10)=50 .

Mathematics
1 answer:
Elina [12.6K]3 years ago
8 0

Step-by-step explanation:

dA/dt = 6 − 0.02A

dA/dt = -0.02 (A − 300)

Separate the variables.

dA / (A − 300) = -0.02 dt

Integrate.

ln(A − 300) = -0.02t + C

Solve for A.

A − 300 = Ce^(-0.02t)

A = 300 + Ce^(-0.02t)

Use initial condition to find C.

50 = 300 + Ce^(-0.02 × 10)

50 = 300 + Ce^(-0.2)

-250 = Ce^(-0.2)

C = -250e^(0.2)

A = 300 − 250e^(0.2) e^(-0.02t)

A = 300 − 250e^(0.2 − 0.02t)

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Let equation be Ax+By+C=0
then -A/B=-√3, or
B=A/√3.
Equation becomes
Ax+(A/√3)y+C=0

Knowing that line is 8 units from origin, apply distance formula
8=abs((Ax+(A/√3)y+C)/sqrt(A^2+(A/√3)^2))
Substitute coordinates of origin (x,y)=(0,0)  =>
8=abs(C/sqrt(A^2+A^2/3))
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solve for positive solution of C
8=C/√(4/3) => C=8*2/√3 = (16/3)√3

Therefore one solution is
x+(1/√3)+(16/3)√3=0
or equivalently
√3 x + y + 16 = 0

Check:
slope = -1/√3  .....ok
distance from origin
= (√3 * 0 + 0 + 16)/(sqrt(√3)^2+1^2)
=16/2
=8  ok.

Similarly C=-16 will satisfy the given conditions.

Answer  The required equations are
√3 x + y = ± 16 
in standard form.

You can conveniently convert to point-slope form if you wish.




4 0
3 years ago
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