Question

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The area a of a blot of ink is growing such that after t second a = 3t² + t/5 + 7 . Calculate the rate of increase of area after 5s . If the motion of a particle is represented by
s = t³+ t² - t + 2 , find the position , velocity and acceleration of the particle after 2s .

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Answer 1

Answer:

Question 1:

Let the area that the inkblot spreads for every t seconds be A.

We'll find the rate of change of the inkblot for every t seconds by differentiating A. We'll be differentiating 'A' with respect to 't'. Once we do so, we'll be able to find the rate of increase in the area after 5 seconds.

$\sf \dashrightarrow A = 3t^2 + \dfrac{t}{5} + 7$

Differentiate on both sides.

$\sf \Longrightarrow \dfrac{d}{dt} \Big(A\Big) = \dfrac{d}{dt} \bigg(3t^2 + \dfrac{t}{5} + 7\bigg)$

Differentiate each addend.

$\sf \Longrightarrow \dfrac{dA}{dt} = \dfrac{d}{dt} \bigg(3t^2\bigg) + \dfrac{d}{dt}\bigg(\dfrac{t}{5}\bigg) + \dfrac{d}{dt}\bigg(7\bigg)$

Take out the constant factors.

$\sf \Longrightarrow \dfrac{dA}{dt} = 3 \cdot \dfrac{d}{dt} \bigg(t^2\bigg) + \bigg(\dfrac{1}{5}\bigg) \cdot \dfrac{d}{dt}\bigg(\dfrac{t}{1}\bigg) + \dfrac{d}{dt}\bigg(7\b$

Derivative of a constant term is equal to 0, ∴ (d/dt)(7) is equal to 0.

$\sf \Longrightarrow \dfrac{dA}{dt} = 3 \cdot \dfrac{d}{dt} \bigg(t^2\bigg) + \bigg(\dfrac{1}{5}\bigg) \cdot \dfrac{d}{dt}\bigg(t\bigg) + \ \textsf{\textbf{0}}$

Derivative of the differentiation variable ('t' in this case) is equal to 1, ∴ (d/dt)(t) is equal to 1.

$\sf \Longrightarrow \dfrac{dA}{dt} = 3 \cdot \dfrac{d}{dt} \bigg(t^2\bigg) + \bigg(\dfrac{1}{5}\bigg) \cdot \big(1\big) + 0$

$\sf \Longrightarrow \dfrac{dA}{dt} = 3 \cdot \dfrac{d}{dt} \bigg(t^2\bigg) + \dfrac{1}{5}$

Acc. to the power rule;

$\sf \dashrightarrow \dfrac{d}{dx} \Big(x^n\Big) = nx^{[n - 1]}$

Here, x = t and n = 2. On using the power rule;

$\sf \Longrightarrow \dfrac{dA}{dt} = 3 \cdot \Big(2t^{[2 - 1]}\Big) + \dfrac{1}{5}$

$\sf \Longrightarrow \dfrac{dA}{dt} = 3 \cdot \Big(2t^{1}\Big) + \dfrac{1}{5}$

$\sf \Longrightarrow \dfrac{dA}{dt} = 3 \cdot \Big(2t\Big) + \dfrac{1}{5}$

$\sf \Longrightarrow \dfrac{dA}{dt} = \textsf{\textbf{6t}} + \dfrac{\textsf{\textbf{1}}}{\textsf{\textbf{5}}}$

Substitute t = 5, as we need to find the rate of increase in area after 5 seconds.

$\sf \Longrightarrow \dfrac{dA}{dt} = 6(5) + \dfrac{1}{5}$

$\sf \Longrightarrow \dfrac{dA}{dt} = 30 + \dfrac{1}{5}$

Take LCM;

$\sf \Longrightarrow \dfrac{dA}{dt} = \dfrac{150 + 1}{5}$

$\sf \Longrightarrow \dfrac{dA}{dt} = \dfrac{151}{5}$

$\sf \Longrightarrow \dfrac{dA}{dt} = \textsf{\textbf{30.2 sq.units/sec}}$

=30.2 sq.units/sec

Answer: 30.2 sq.units/sec.

Question 2:

We've been given the position of a particle as 's' (displacement); s = t³ + t² - t + 2.

$\sf \Longrightarrow s = t^3 + t^2 - t + 2$

Substitute t = 2.

$\sf \Longrightarrow s = (2)^3 + (2)^2 - (2) + 2$

$\sf \Longrightarrow s = 8 + 4 - 2 + 2$

$\sf \Longrightarrow \textsf{\textbf{s = 12m}}$

We know that;

$\sf \Longrightarrow Velocity (v) = \dfrac{ds}{dt}$

On substituting the value of 's' we get; ['s' is displacement, 't' is time taken]

$\sf \Longrightarrow v = \dfrac{d\Big(t^3 + t^2 - t + 2\Big)}{dt}$

$\sf \Longrightarrow v = \dfrac{d}{dt}\Big(t^3\Big) + \dfrac{d}{dt}\Big(t^2\Big) - \dfrac{d}{dt}\Big(t\Big) + \dfrac{d}{dt}\Big(2\Big)$

Wkt;

Derivative of a constant term is equal to 0.

Derivative of the differentiation variable is equal to 1.

On applying these relations we get;

$\sf \Longrightarrow v = \dfrac{d}{dt}\Big(t^3\Big) + \dfrac{d}{dt}\Big(t^2\Big) - \textsf{\textbf{1 + 0}}$

On using the power rule we get;

$\sf \Longrightarrow v =\Big(3t^{[3 - 1]}\Big) + \Big(2t^{[2 - 1]}\Big) - 1$

$\sf \Longrightarrow v =\Big(3t^{2}\Big) + \Big(2t\Big) - 1$

$\sf \Longrightarrow v =\textsf{\textbf{3t}}^{2} + \textsf{\textbf{2t - 1}}$

On substituting t = 2 seconds we get;

$\sf \Longrightarrow v = 3(2)^{2} + 2(2) - 1$

$\sf \Longrightarrow v = 3(4) + 4 - 1$

$\sf \Longrightarrow v = 12 + 3$

$\sf \Longrightarrow \textsf{\textbf{v = 15 m/s}}$

We know that;

$\sf \Longrightarrow Acceleration (a) = \dfrac{dv}{dt}$

Substitute v = 3t² + 2t - 1.

$\sf \Longrightarrow a = \dfrac{d(3t^{2} + 2t - 1)}{dt}$

$\sf \Longrightarrow a = \dfrac{d}{dt}(3t^{2}) + \dfrac{d}{dt}(2t) - \dfrac{d}{dt}(1)$

$\sf \Longrightarrow a = 3 \cdot \dfrac{d}{dt}(t^{2}) + \ 2 \cdot \dfrac{d}{dt}(t) - \dfrac{d}{dt}(1)$

Walt;

Derivative of a constant term is equal to 0.

Derivative of the differentiation variable is equal to 1.

On applying these relations and the power rule we get;

$\sf \Longrightarrow a = 3(2 \times t)^{[2 - 1]} + 2(1) - 0$

$\sf \Longrightarrow a = 3(2t)^{1} + 2(1) - 0$

$\sf \Longrightarrow a = 6t + 2$

Substitute t = 2 seconds.

$\sf \Longrightarrow a = 6(2) + 2$

$\sf \Longrightarrow a = 12 + 2$

$\sf \Longrightarrow \textsf{\textbf{a = 14 m/s}}\ ^{2}$

Therefore;

• Position = 12m
• Velocity = 15 m/s
• Acceleration = 14 m/s²