Answer to Question #158454 in Calculus for geek

Question

Find the equation of the line passes through the point (3, −2) and is perpendicular to the line 3x − 2y = 4.

An environmental study of a certain community suggests that the average daily level of pollution in the air will be Q(p) = √ 0.6p + 20 units when the population is p thousand. It is estimated that after t years the population will be p(t) = 9 + 0.5t²Â  thousand.

(a) Express the level of pollution in the air as a function of time.

(b) compute the level of pollution after 5 years from now.

(c) When will the pollution level reach 10 units? 

Answer 1

Solution 1:

Let the slope of the required line be $m_1$

Given line: $3x - 2y = 4$

$\Rightarrow 3x - 4 = 2y \\ \Rightarrow y=\dfrac{3x}2-2$

On comparing with $y=mx+c$ , we get,

$m=\dfrac32$ , i.e. slope of given line.

Since, required line is perpendicular to this given line, product of their slopes is -1.

$m\times m_1=-1 \\ \Rightarrow \dfrac32 \times m_1=-1 \\ \Rightarrow m_1=-1 \times \dfrac23=\dfrac{-2}3$

Also, given point is (3, -2), i.e. $x_1=3,\ y_1=-2$

Now, equation of line with slope and a point is:

$y-y_1=m_1(x-x_1)$

Putting values

$y-(-2)=\dfrac{-2}3(x-3)$

$\Rightarrow y+2=\dfrac{-2x}3+2$

$\Rightarrow y=\dfrac{-2x}3$

$\Rightarrow 2x+3y=0$

Answer: $2x+3y=0$ .

Solution 2:

Given, Q(p)$=\sqrt{0.6p+20}$

And p(t)$=9+0.5t^2$

(a): Q(p) is level of pollution and function of p and p(t) is population after t years.

We need to express Q(p) in terms of t, i.e. composition function, $Q \omicron P(t)$ .

$Q \omicron P(t)=Q[P(t)]=\sqrt{0.6(9+0.5t^2)+20}$

$=\sqrt{5.4+0.3t^2+20}$

$=\sqrt{25.4+0.3t^2}$

(b): Put t=5

$Q \omicron P(5)=\sqrt{25.4+0.3(5)^2}$

$=\sqrt{25.4+7.5}$

$=\sqrt{32.9}$

$\approx 5.74$ units

(c): Now, $Q \omicron P(t)=10$ units

$\sqrt{25.4+0.3t^2}=10$

Now solving for t,

${25.4+0.3t^2}=100$ [On squaring both sides]

$\Rightarrow {0.3t^2}=100-25.4=74.6$

$\Rightarrow t^2=\dfrac{74.6}{0.3}\approx248.66$

$\Rightarrow t=\sqrt{248.66}=15.76$

Answer:

(a) $\sqrt{25.4+0.3t^2}$

(b) 5.74 units

(c) 15.76 years.