Find, to the nearest °C, the temperature the coffee has reached when $T'(t) = -4.05$, where $T'(t)$ is the rate at which the coffee is cooling, in °C per minute. A sugar cube is put into the coffee. The sugar keeps its cube shape as it dissolves. As the sugar cube dissolves, its volume decreases at the constant rate of $\frac{1}{20} \, cm^3/sec$. Let $x(t)$ be the sidelength of the cube at time $t$. Find the rate of change of $x(t)$ when the volume of the cube reaches $\frac{64}{3} \, cm^3$.

Leaving Cert Mathematics Differentiation: Find, to the nearest °C, the tem

Find, to the nearest °C, the temperature the coffee has reached when $T'(t) = -4.05$, where $T'(t)$ is the rate at which the coffee is cooling, in °C per minute - Leaving Cert Mathematics - Question 9 - 2021

Question 9

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Find, to the nearest °C, the temperature the coffee has reached when $T'(t) = -4.05$, where $T'(t)$ is the rate at which the coffee is cooling, in °C per minute. A ... show full transcript

Worked Solution & Example Answer:Find, to the nearest °C, the temperature the coffee has reached when $T'(t) = -4.05$, where $T'(t)$ is the rate at which the coffee is cooling, in °C per minute - Leaving Cert Mathematics - Question 9 - 2021

Step 1

Find Temperature when $T'(t) = -4.05$

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Answer

Given the formula for temperature:
$T(t) = 75e^{-0.08t} + 20$
To find the temperature when $T'(t) = -4.05$ , we first calculate $T'(t)$ :
$T'(t) = -75(0.08)e^{-0.08t}$
Setting $T'(t) = -4.05$ gives:
$-75(0.08)e^{-0.08t} = -4.05$
Solving for $e^{-0.08t}$ :
$e^{-0.08t} = \frac{4.05}{6} = \frac{2}{5}$
Taking the natural logarithm of both sides:
$-0.08t = ext{ln}\left(\frac{2}{5}\right)$
Thus,
$t = \frac{-\text{ln}\left(\frac{2}{5}\right)}{0.08} \approx 5.007$
Now substituting $t$ back into the temperature equation:
$T(5.007) = 75e^{-0.08(5.007)} + 20 \approx 70 \degree C$

Step 2

Find the rate of change of $x(t)$

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Answer

The volume of the cube, $V$ , is given by the formula:
$V = x(t)^3$
Differentiating with respect to $t$ :
$\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$
We know that the volume dissolves at a constant rate of $\frac{1}{20} \, cm^3/sec$ , so
$\frac{dV}{dt} = -\frac{1}{20}$
Setting this equal:
$-\frac{1}{20} = 3x^2 \frac{dx}{dt} \implies \frac{dx}{dt} = -\frac{1}{20 \cdot 3x^2}$
At the point where the volume reaches $\frac{64}{3} \, cm^3$ :
$\frac{64}{3} = x^3 \implies x = \left(\frac{64}{3}\right)^{\frac{1}{3}} \approx 2.88$
Substituting back:
$\frac{dx}{dt} = -\frac{1}{20 \cdot 3(2.88)^2} \approx -\frac{1}{20 \cdot 3 \cdot 8.26} \approx -\frac{1}{496.8} \approx -0.00201 \, cm/sec$

Find, to the nearest °C, the temperature the coffee has reached when $T'(t) = -4.05$, where $T'(t)$ is the rate at which the coffee is cooling, in °C per minute - Leaving Cert Mathematics - Question 9 - 2021

Worked Solution & Example Answer:Find, to the nearest °C, the temperature the coffee has reached when $T'(t) = -4.05$, where $T'(t)$ is the rate at which the coffee is cooling, in °C per minute - Leaving Cert Mathematics - Question 9 - 2021

Find Temperature when $T'(t) = -4.05$

Find the rate of change of $x(t)$

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