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Question 120 of 126

Q.(a) The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours, find how many bacteria will be present after 10 hours ? OR

(b) Find the vector and Cartesian equations of the plane containing x−22=y−23=z−13\dfrac{x-2}{2}=\dfrac{y-2}{3}=\dfrac{z-1}{3} and parallel to the line x+13=y−12=z+11\dfrac{x+1}{3}=\dfrac{y-1}{2}=\dfrac{z+1}{1}.
Puducherry TnboardTamil Nadu HSC (DGE) Board 2024Subjective· 5mImportance★★★★★
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(a) Solves the exponential-growth model dP/dt=kPdP/dt=kP using the given tripling time to predict the count at double that time; (b) builds the required plane from a point on the given line, the line's own direction, and the direction of the line it must stay parallel to. Both alternatives answered below.

(a) Exponential bacterial growth

1. Model. "Rate of increase proportional to the number present" is dPdt=kP\dfrac{dP}{dt}=kP, whose solution is P(t)=P0ektP(t)=P_0e^{kt}, where P0P_0 is the initial count.

2. Use the tripling condition. Given P(5)=3P0P(5)=3P_0: P0e5k=3P0⇒e5k=3P_0e^{5k}=3P_0\Rightarrow e^{5k}=3.

3. Find P(10)P(10).

P(10)=P0e10k=P0(e5k)2=P0(3)2=9P0P(10)=P_0e^{10k}=P_0\left(e^{5k}\right)^2=P_0(3)^2=9P_0

4. So after 1010 hours, the bacteria count is 99 times the initial number (it triples again over the second 5-hour interval).

(b) Plane containing one line, parallel to another

1. Data. The plane contains the line x−22=y−23=z−13\dfrac{x-2}2=\dfrac{y-2}3=\dfrac{z-1}3, so it passes through A(2,2,1)A(2,2,1) with direction d⃗1=(2,3,3)\vec d_1=(2,3,3). It is parallel to the line x+13=y−12=z+11\dfrac{x+1}3=\dfrac{y-1}2=\dfrac{z+1}1, with direction d⃗2=(3,2,1)\vec d_2=(3,2,1) — this direction must also lie in the plane.

2. Normal of the plane. …

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