Answer:
a)
And since the power can't be negative then the solution would be w = 33.34 watts.
b) ![202= 0.1w^2 +2.156 w +20](https://tex.z-dn.net/?f=%20202%3D%200.1w%5E2%20%2B2.156%20w%20%2B20)
![204= 0.1w^2 +2.156 w +20](https://tex.z-dn.net/?f=%20204%3D%200.1w%5E2%20%2B2.156%20w%20%2B20)
![0.1w^2 +2.156 w-184=](https://tex.z-dn.net/?f=%200.1w%5E2%20%2B2.156%20w-184%3D)
So then the range of voltage would be between 33.22 W and 33.45 W.
c)For this case
since that's the tolerance 1C
d) ![\delta_1 =|33.222-33.336|=0.116](https://tex.z-dn.net/?f=%20%5Cdelta_1%20%3D%7C33.222-33.336%7C%3D0.116)
![\delta_2 =|33.449-33.336|=0.113](https://tex.z-dn.net/?f=%20%5Cdelta_2%20%3D%7C33.449-33.336%7C%3D0.113)
So then we select the smalles value on this case ![\delta =0.113](https://tex.z-dn.net/?f=%20%5Cdelta%20%3D0.113)
e) For this case if we assume a tolerance of
for the temperature and a tolerance for the power input
we see that:
![lim_{x \to a} f(x) =L](https://tex.z-dn.net/?f=%20lim_%7Bx%20%5Cto%20a%7D%20f%28x%29%20%3DL)
Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C
Step-by-step explanation:
For this case we have the following function
![T(w)= 0.1 w^2 +2.156 w +20](https://tex.z-dn.net/?f=%20T%28w%29%3D%200.1%20w%5E2%20%2B2.156%20w%20%2B20)
Where T represent the temperature in Celsius and w the power input in watts.
Part a
For this case we need to find the value of w that makes the temperature 203C, so we can set the following equation:
![203= 0.1w^2 +2.156 w +20](https://tex.z-dn.net/?f=%20203%3D%200.1w%5E2%20%2B2.156%20w%20%2B20)
And we can rewrite the expression like this:
![0.1w^2 +2.156 w-183=](https://tex.z-dn.net/?f=%200.1w%5E2%20%2B2.156%20w-183%3D)
And we can solve this using the quadratic formula given by:
![w =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}](https://tex.z-dn.net/?f=%20w%20%3D%5Cfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E2%20-4ac%7D%7D%7B2a%7D)
Where a =0.1, b =2.156 and c=-183. If we replace we got:
And since the power can't be negative then the solution would be w = 33.34 watts.
Part b
For this case we can find the values of w for the temperatures 203-1= 202C and 203+1 = 204 C. And we got this:
![202= 0.1w^2 +2.156 w +20](https://tex.z-dn.net/?f=%20202%3D%200.1w%5E2%20%2B2.156%20w%20%2B20)
![0.1w^2 +2.156 w-182=](https://tex.z-dn.net/?f=%200.1w%5E2%20%2B2.156%20w-182%3D)
![204= 0.1w^2 +2.156 w +20](https://tex.z-dn.net/?f=%20204%3D%200.1w%5E2%20%2B2.156%20w%20%2B20)
![0.1w^2 +2.156 w-184=](https://tex.z-dn.net/?f=%200.1w%5E2%20%2B2.156%20w-184%3D)
So then the range of voltage would be between 33.22 W and 33.45 W.
Part c
For this case
since that's the tolerance 1C
Part d
For this case we can do this:
![\delta_1 =|33.222-33.336|=0.116](https://tex.z-dn.net/?f=%20%5Cdelta_1%20%3D%7C33.222-33.336%7C%3D0.116)
![\delta_2 =|33.449-33.336|=0.113](https://tex.z-dn.net/?f=%20%5Cdelta_2%20%3D%7C33.449-33.336%7C%3D0.113)
So then we select the smallest value on this case ![\delta =0.113](https://tex.z-dn.net/?f=%20%5Cdelta%20%3D0.113)
Part e
For this case if we assume a tolerance of
for the temperature and a tolerance for the power input
we see that:
![lim_{x \to a} f(x) =L](https://tex.z-dn.net/?f=%20lim_%7Bx%20%5Cto%20a%7D%20f%28x%29%20%3DL)
Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C