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