(a) A police jeep, approaching an orthogonal intersection from the northern direction, is chasing a speeding c

Question 122 of 148

Q. (a) A police jeep, approaching an orthogonal intersection from the northern direction, is chasing a speeding car that has turned and moving straight east. When the jeep is 0.6 km north of the intersection and the car is 0.8 km to the east, the police determine with a radar that the distance between the jeep and the car is increasing at 20 km/hr. If the jeep is moving at 60 km/hr at the instant of measurement, what is the speed of the car ? OR

(b) Find the area of the region bounded by

x

-axis, the curve

y=|\cos x|

, the lines

x=0

and

x=\pi

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2020Subjective· 5mImportance★★★★★

82% · 122/148 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Meaning of Derivative and Rate of Change

The derivative $f'(x)$ of a function carries two equivalent readings, and this chapter leans on both throughout.

As a slope. For the curve $y=f(x)$ , the slope of the chord joining $(x,f(x))$ and $(x+h,f(x+h))$ is the Newton quotient

\frac{f(x+h)-f(x)}{h}.

Taking $h\to0$ gives the slope of the curve at $(x,f(x))$ itself:

f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.

If $\theta$ is the angle the tangent makes with the positive $x$ -axis (measured anticlockwise), then $f'(x)=\tan\theta$ .

As a rate of change. $f'(x)=\dfrac{dy}{dx}$ is also the instantaneous rate of change of $y$ with respect to $x$ ; over an interval $[a,b]$ the average rate of change is the ordinary difference quotient $\dfrac{f(b)-f(a)}{b-a}$ (a chord slope), while the derivative at a single point is the instantaneous rate.

Motion along a line. If $s=f(t)$ is the position of an object at time $t$ (measured from a fixed origin, positive direction = forward):

v(t)=\frac{ds}{dt},\qquad a(t)=\frac{dv}{dt}=\frac{d^2s}{dt^2}.

Speed $=|v(t)|=\left|\dfrac{ds}{dt}\right|$ — always non-negative, regardless of direction.
$v(t)=0$ : the particle is momentarily at rest.
$v(t)>0$ : moving forward; $v(t)<0$ : moving backward.
The particle changes direction exactly where $v(t)$ changes sign (not merely where it is zero — the sign must flip on either side).
If the particle reverses direction at time $t_c\in(t_1,t_2)$ , the total distance travelled from $t_1$ to $t_2$ is $|s(t_c)-s(t_1)|+|s(t_2)-s(t_c)|$ — NOT simply $|s(t_2)-s(t_1)|$ , since backtracking would otherwise cancel out.
Near Earth's surface a freely falling body has constant acceleration $g\approx9.8\ \text{m/s}^2$ ( $32\ \text{ft/s}^2$ ), giving $a=-g,\ v=-gt+v_0,\ s=-\tfrac12gt^2+v_0t+s_0$ . …

Start your 14-day free trial to unlock the full solution →

(a) is a related-rates problem: differentiate $z^2=x^2+y^2$ (distance between the jeep on the north axis and the car on the east axis) with respect to time and solve for the car's speed; (b) integrates $|\cos x|$ over $[0,\pi]$ by splitting at the sign change at $x=\pi/2$ .

(a) Speed of the car

Put the intersection at the origin $O$ . The jeep travels along the north axis, position $(0,y)$ , moving toward $O$ so $y$ is decreasing: $\dfrac{dy}{dt}=-60$ km/hr (jeep speed $60$ km/hr). The car travels along the east axis, position $(x,0)$ , moving away from $O$ : $\dfrac{dx}{dt}$ is the unknown car speed (positive).
Distance between them: $z^2=x^2+y^2$ . Differentiate w.r.t. $t$ : $2z\dfrac{dz}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}\Rightarrow z\dfrac{dz}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$ .
At the instant given: $x=0.8$ , $y=0.6$ , so $z=\sqrt{0.8^2+0.6^2}=\sqrt{0.64+0.36}=\sqrt1=1$ km. Also $\dfrac{dz}{dt}=20$ km/hr (distance increasing).
Substitute: $1(20)=0.8\dfrac{dx}{dt}+0.6(-60)\Rightarrow20=0.8\dfrac{dx}{dt}-36\Rightarrow0.8\dfrac{dx}{dt}=56\Rightarrow\dfrac{dx}{dt}=70$ .
The car's speed is $70$ km/hr. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2020Set ANNUAL1 markMCQ
$s(t) = 3t^2 - 2t - 8$
(a) $t=3$
(b) $t=0$
(c) $t=\dfrac{1}{3}$
(d) $t=1$
›Reveal solutionSolution
Setting the velocity $v(t)=\dfrac{ds}{dt}$ to zero gives the time at which the particle is at rest, $t=\dfrac13$ .

The position function is $s(t)=3t^2-2t-8$ .

The velocity is the first derivative of position with respect to time: $v(t)=\dfrac{ds}{dt}=6t-2$ . …
Start your 14-day free trial to see the full solution →
CBSE 2019Set ANNUAL1 markMCQ
$Q. The surface area of a sphere when the volume is increasing at the same rate as its radius, is :$
(a) $4\pi$
(b) $\dfrac{4\pi}{3}$
(c) $1$
(d) $\dfrac{1}{2\pi}$
›Reveal solutionSolution
When the rate of change of volume equals the rate of change of radius, the sphere's surface area equals $1$ .

Volume of a sphere: $V=\dfrac{4}{3}\pi r^3$ .

Differentiate with respect to time $t$ : $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ .

We are given $\dfrac{dV}{dt}=\dfrac{dr}{dt}$ (the volume increases at the same rate as the radius).

Substitute: $\dfrac{dr}{dt}=4\pi r^2\dfrac{dr}{dt}$ . …
Start your 14-day free trial to see the full solution →
CBSE 2017Set ANNUAL1 markMCQ
$y = F(t)$
(a) Gradient of the velocity/time graph
(b) Gradient of the distance/time graph
(c) Gradient of the acceleration/time graph
(d) Gradient of the velocity/distance graph
›Reveal solutionSolution
Since velocity is $dy/dt$ and acceleration is $dv/dt$ , acceleration is the gradient of the velocity–time graph.

Given $y=F(t)$ is displacement (distance) as a function of time $t$ .

Velocity $v=\dfrac{dy}{dt}$ is, by definition, the gradient (slope) of the distance–time graph.

Acceleration $a=\dfrac{dv}{dt}$ is the rate of change of velocity with respect to time — i.e., the gradient of the velocity–time graph. …
Start your 14-day free trial to see the full solution →

←Previous questionQ36 — Find the critical numbers (only x values) of the function f(x)=x4/5(x-…→Next questionQ38 — (a) A square shaped thin material with area 196 sq. units to make into…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days