Question

T= 300 (d−15) 2 ​ +20space, T, equals, start fraction, left parenthesis, d, minus, 15, right parenthesis, start superscript, 2, end superscript, divided by, 300, end fraction, plus, 20 The average temperature around the Tropic of Cancer for a particular day in January can be modeled using the equation above, where TTT is the temperature in degrees Celsius and ddd is the number of days after January 111 (on January 1, d=01,d=01, comma, d, equals, 0). According to this model, what is the lowest average temperature, in degrees Celsius, for a day in January, which has 313131 days?

Answer 1

Answer:

R(t)=-0.9sin(2π/365t)+2.3

Step-by-step explanation:

Answer 2

As can be read from your statement written "T, equals, start fraction, left parenthesis, d, minus, 15, right parenthesis, strt superscript, 2, end superscript, divided by, 300, end fraction, plus, 20", I hope your model equation is this :

$T = \frac{(d-15)^{2}}{300} + 20$

Hope this is your question, if not I think you will, still be able to find an answer of your question based on this solution

As we have to find lowest average temperature, So for minimum of a function its derivative is equal to 0 there.

So lets find derivative of T function first

So first expand $(d-15)^{2}$ as (d-15)(d-15)

we will use FOIL to multiply these

so $(d-15)(d-15) = d^{2} -15d -15d +225$

$= d^{2} -30d +225$

so we have $T = \frac{d^{2} -30d +225}{300} +20$

Now we will derivate each term here,300 in denominator is constant so that will come as it in in denominator.

To derivate terms in $dx^{2} -30d +225$ we will use power rule formula:

$(x^{n} )'= nx^{n-1}$

so derivative of $(d^{2} )'= 2d^{2-1} = 2d^{1} = 2d$

Then derivative of d will be 1

so that of -30d will be -30

then derivate of constant -225 will be 0

so we will have derivative as $\frac{2d-30}{300}$ for the fraction part and then derivative of +20 is again 0 as its constant term

$T' = \frac{2d-30}{300}$

For minimum we will put this derivative =0

$0 = \frac{2d-30}{300}$

Now solve for d

times both sides by 300

$0 \times 300 = \frac{2d-30}{300} \times 300$

$0 = 2d-30$

$0 +30 = 2d -30 +30$

$30 = 2d$

$\frac{30}{2} = \frac{2d}{2}$

$15 = d$

So now we have to find value of lowest temperature.

For that simply plug 15 in d place in original T function equation

$T = \frac{(d-15)^{2}}{300} + 20$

$T = \frac{(15-15)^{2}}{300} + 20$

$T = 20$

So T = 20 °C is the lowest average temperature and the answer.