Question

(a) A bowl of water has a temperature of 50◦C. It is put into a refrigerator where the temper-

ature is 5◦C. After 0.5 hours, the water is stirred and its temperature is measured to be

20◦C. It is then left to cool further, take t in Newton’s law of cooling to be measured in

minutes. Use Newton’s law of cooling to predict when the temperature will be 10◦C.


(b) Verify that the given family of functions solves the differential equation.

(i) dy

dt = (1 − 2t)y

2

, y =

1

C − t + t

2

.

(ii) dy

dt = y

2

sin t, y =

1

C + cost

.

(c) Evaluate the integral Z 1

0

x

2

(

√

4 − x

2)

3

dx

Answer 1

(a) Equation of newton's law of cooling,

$\int \frac{dT}{(T-T_0)} = - \int k dt$

Where T is temperature at any time t and T₀ is the temperature of surrounding.

Given, (i) at t = 0, T = 50⁰C = 323K

(ii) at t = 0.5 hrs, T = 20 C = 293 K

Temperature of the surrounding is T₀ = 5 C = 293 K

Solving equation we obtain,

$ln(T-T_0) = -kt + lnC$

applying condition (i)

$lnC = ln (45)$

applying (ii) condition,

$k = 2ln(3)$

Hence final equation will be,

$ln(T-293) = -2ln(3) t + ln(45)$

If T = 10 C = 283 K, then for time calculation,

$ln(283-278) = -2ln(3)t + ln(45) \implies t = 1 hour$

b (i) Given $\frac{dy}{dt} = (1-2t)y^2$ and $y = \frac{1}{c-t+t^2}$

Differentiating y with respect to x, 

$\frac{dy}{dx} = -(\frac{1}{c-t+t^2})^2(-1+2t) = (\frac{1}{c-t+t^2})^2(1-2t)$

replacing value of y 

$\frac{dy}{dt} = (1-2t)y^2$

(ii) $\frac{dy}{dt} = y^2sin(t)$ and $y = \frac{1}{c+cost}$

differentiating both sides with respect to x,

$\frac{dy}{dt} = -(\frac{1}{c+cost})^2(-sint) = (\frac{1}{c+cost})^2(sint) = y^2sin(t)$

(iii) $\int_0^1 x^2 (\sqrt{4-x^2})^3dx$

let $x = 2sin(u)$ then $dx = 2cos(u) du$

Let us solve integration first without limit

$I=\int 8cos(u)sin^2(u)(4-4sin^2(u))^{3/2}du =\int 8cos(u)sin^2(u)(4cos^2(u))^{3/2}du =\int 64cos^4(u)sin^2(u)du = \int 64cos^6(u)(1-cos^2(u))du$

Integrating it,

$I = -\frac{x(4-x^2)^{5/2}}{6} + \frac{x(4-x^2)^{3/2}}{6}+x(4-x^2)^{1/2} + 4asin(\frac{x}{2})$

putting limits,

$\int_0^1 x^2 (\sqrt{4-x^2})^3dx = \frac{2\pi}{3}$