Informative line

### Differentiation And Its Meaning

Learn slope of tangent and secant line, average and instantaneous velocity formula calculus. Practice to find the slope of the curve at the given point.

# The Tangent Problem

Consider a curve $$C$$, and two points  $$A(a,\,f(a))$$  and $$B(x,\,f(x))$$ on it, the curve being  $$y=f(x)$$  slop of secant line AB $$\dfrac{f(x)-f(a)}{x-a}$$  slope of secant line AB = $$\dfrac{f(b)-(f(a)}{b-a}$$

• Slope of line joining $$(x_1,\,y_1)$$ and $$(x_2,\,y_2)$$ is

$$m=\dfrac{y_2–y_1}{x_2–x_1}$$  slope of tangent at A= m$$\lim\limits_{x\to a} \dfrac{f(x)-f(a)}{x-a}$$

As B approaches A, this secant AB will become tangent at A.  We are assuming that the above limit exists.

#### The slope of tangent line to the curve $$y=x^2$$ at the point $$P(2,\,4)$$ on it, is

A –7

B 4

C 0

D 42

×

$$m_t=\lim\limits_{x\to a}\left(\dfrac{f(x)–f(a)}{x–a}\right)$$

$$a=2$$,

$$=\lim\limits_{x\to 2}\;\dfrac{f(x)–f(2)}{x–2}$$

$$=\lim\limits_{x\to 2}\;\dfrac{x^2–4}{x–2}$$

$$=\lim\limits_{x\to 2}\;\dfrac{(x–2)(x+2)}{x–2}$$

$$=4$$ $$m_t=4$$

### The slope of tangent line to the curve $$y=x^2$$ at the point $$P(2,\,4)$$ on it, is

A

–7

.

B

4

C

0

D

42

Option B is Correct

# Equation of a Line

The equation of a straight line which passes through a fixed point $$(x_1,\,y_1)$$ and whose slope is $$m$$ is given by,

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

$$m=tan\theta=\dfrac{y-y_1}{x-x_1}$$  • A line is completely determined if its slope and a fixed point is given on it.

#### The equation of line which passes through $$(-2,\,7)$$ and whose slope is 2, is

A $$x+y+8=0$$

B $$2x-y+11=0$$

C $$2x=3$$

D $$y=7$$

×

Equation of line $$\to\,y-y_1=m(x-x_1)$$

$$\Rightarrow\,y-7=2(x-(-2))$$

$$\Rightarrow\,y-7=2(x+2)$$

$$\Rightarrow\,y-7=2x+4$$

$$2x-y+11=0$$

### The equation of line which passes through $$(-2,\,7)$$ and whose slope is 2, is

A

$$x+y+8=0$$

.

B

$$2x-y+11=0$$

C

$$2x=3$$

D

$$y=7$$

Option B is Correct

# Solving Problems related to Velocity

A particle moves in a straight line with displacement function, $$s=f(t)$$ at the time $$'t'$$ then

• $$\text{Average velocity}=\dfrac{f(t_2)-f(t_1)}{t_2-t_1}=\dfrac{\text{displacement}}{\text{time}}$$

$$=\dfrac{f(a+h)-f(a)}{h}\to (t_1=a\,\text{ and}\,\,t_2=a+h)$$

$$\Rightarrow\,\text{Average velocity}=\dfrac{f(a+h)-f(a)}{h}$$

#### The displacement of a particle $$'s'$$ in  meters is given by  $$s=f(t)=t^3-2t^2+1$$  where $$'t'$$ is in seconds.  Find the average velocity in the time interval [2, 4].

A $$78\,m/s$$

B $$-2m/s$$

C $$0.5m/s$$

D $$16m/s$$

×

$$\text{Average velocity}=\dfrac{f(a+h)-f(a)}{h}$$

$$a=2,\;h=2$$

$$Average\,velocity=\dfrac{f(4)-f(2)}{4-2}$$

$$=\dfrac{(4^3-2×4^2+1)-(2^3-2×2^2+1)}{4-2}$$

$$=\dfrac{(64-32+1)-(8-8+1)}{2}$$

$$=\dfrac{32}{2}$$

$$=16m/s$$

### The displacement of a particle $$'s'$$ in  meters is given by  $$s=f(t)=t^3-2t^2+1$$  where $$'t'$$ is in seconds.  Find the average velocity in the time interval [2, 4].

A

$$78\,m/s$$

.

B

$$-2m/s$$

C

$$0.5m/s$$

D

$$16m/s$$

Option D is Correct

# Instantaneous Velocity

The instantaneous velocity $$v(a)$$ at the time  $$t=a$$  is defined as limit of average velocity.

$$v(a)=\lim\limits_{h\to 0}\;\left(\dfrac{f(a+h)-f(a)}{h}\right)$$

where, $$f(t)$$ = displacement at time $$'t'$$.  • Instantaneous velocity will be simply referred to as velocity at  $$t=a$$.

#### The displacement of a particle moving in a straight line is given by $$s=f(t)=2t^2+t+3$$  where '$$s$$' is in meters and $$'t'$$ is in seconds. The instantaneous velocity at $$t=2$$  is

A $$-7\,m/s$$

B $$9\,m/s$$

C $$12\,m/s$$

D $$18\,m/s$$

×

$$\text{Instantaneous velocity}=v(a)=\lim\limits_{h\to 0}\;\dfrac{f(a+h)-f(a)}{h}$$

$$v(2)=\lim\limits_{h\to 0}\;\left(\dfrac{f(2+h)-f(2)}{h}\right)$$

$$=\lim\limits_{h\to 0}\;\dfrac{\left(2(2+h)^2+(2+h)+3\right)-(2×4+2+3)}{h}$$

$$=\lim\limits_{h\to 0}\;\dfrac{2(4+h^2+4h)+2+h+3-13)}{h}$$

$$=\lim\limits_{h\to 0}\;\dfrac{2h^2+9h}{h}$$

$$=\lim\limits_{h\to 0}\;2h+9$$

$$=9$$

$$v(2)=9\,m/s$$

### The displacement of a particle moving in a straight line is given by $$s=f(t)=2t^2+t+3$$  where '$$s$$' is in meters and $$'t'$$ is in seconds. The instantaneous velocity at $$t=2$$  is

A

$$-7\,m/s$$

.

B

$$9\,m/s$$

C

$$12\,m/s$$

D

$$18\,m/s$$

Option B is Correct

# Slope of Tangent to a Curve at a Point

The slope of tangent to a curve  $$y=f(x)$$ at the point  $$A(a,\,f(a))$$  is

$$m_t=\lim\limits_{x\to a}\,\left(\dfrac{f(x)-f(a)}{x-a}\right)$$  #### Find the slope of tangent to the curve  $$y=2x^2+7$$ at $$(1,\,7)$$ on it.

A 4

B –4

C 5

D $$\dfrac{1}{5}$$

×

$$m_t=\lim\limits_{x\to a}\,\dfrac{f(x)-f(a)}{x-a}$$

$$\therefore\,m_t=\lim\limits_{x\to 1}\,\left(\dfrac{f(x)-f(1)}{x-1}\right)$$

$$=\lim\limits_{x\to 1}\,\dfrac{(2x^2+7)-(2+7)}{x-1}$$

$$=\lim\limits_{x\to 1}\,\dfrac{2x^2-2}{x-1}$$

$$=\lim\limits_{x\to 1}\,\dfrac{2[(x-1)(x+1)]}{x-1}$$

$$=\lim\limits_{x\to 1}2(x+1)$$

$$=4$$

### Find the slope of tangent to the curve  $$y=2x^2+7$$ at $$(1,\,7)$$ on it.

A

4

.

B

–4

C

5

D

$$\dfrac{1}{5}$$

Option A is Correct

# Different Notation of Slope

$$m_t=\lim\limits_{x\to a}\,\left(\dfrac{f(x)-f(a)}{x-a}\right)$$

Let,  $$x-a=h$$

$$\therefore$$ as $$x\to a$$

$$\Rightarrow\,h\to 0$$

$$\Rightarrow\,m_t=\lim\limits_{h\to 0}\,\dfrac{f(a+h)-f(a)}{h}$$

• This limit will be the slope of tangent only if it exists.

#### The equation of tangent to the curve $$y=\dfrac{3x+2}{x+4}$$ at the point $$(1,\,1)$$ on it, is

A $$y=-3$$

B $$x=8$$

C $$x+2y-1=0$$

D $$2x-5y+3=0$$

×

Equation of tangent $$\to y-y_1=m_t(x-x_1)$$

$$m_t=\lim\limits_{h\to 0}\,\left(\dfrac{f(a+h)-f(a)}{h}\right)$$

$$=\lim\limits_{h\to 0}\,\dfrac{f(1+h)-f(1)}{h}$$

$$=\lim\limits_{h\to 0}\, \dfrac {\left ({\dfrac{3(1+h)+2}{1+h+4}-1}\right)}{h}$$

$$=\lim\limits_{h\to 0}\,\dfrac{3+3h+2-h-5}{h(h+5)}$$

$$\Rightarrow\,m_t=\lim\limits_{h\to 0}\,\dfrac{2h}{h(h+5)}$$

$$=\dfrac{2}{5}$$

Equation $$\to y-1=\dfrac{2}5{}(x-1)$$

$$\Rightarrow\,5y-5=2x-2$$

$$\Rightarrow\,2x-5y+3=0$$

### The equation of tangent to the curve $$y=\dfrac{3x+2}{x+4}$$ at the point $$(1,\,1)$$ on it, is

A

$$y=-3$$

.

B

$$x=8$$

C

$$x+2y-1=0$$

D

$$2x-5y+3=0$$

Option D is Correct