Question

Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x4 − 6x3 + 13x2 − 10x + 5 at (0, 5)
(ii) y = x4 − 6x3 + 13x2 − 10x + 5 at (1, 3)
(iii) y = x3 at (1, 1)
(iv) y = x2 at (0, 0)
(v) x = cos t, y = sin t at t=$\frac{π}{4}$

Answer 1

(i) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5

On differentiating with respect to x, we get:

$\frac{dy}{dx}$ = 4x3-18x2+26x-10

$(\frac{dy}{dx}) \bigg]_{(0,5)}$=-10

Thus, the slope of the tangent at (0, 5) is −10. The equation of the tangent is given as: y − 5 = − 10(x − 0)

⇒ y − 5 = − 10x

⇒ 10x + y = 5

The slope of the normal at (0, 5) is $\frac{-1}{\text{slop of the tangent at}\,(0,5)}$= $\frac{1}{10}.$

Therefore, the equation of the normal at (0, 5) is given as:

y-5=$\frac{1}{10}.$(x-0)

=10y-50=x

x=10y+50=0

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(ii) The equation of the curve is y =x4 − 6x3 + 13x2 − 10x + 5

On differentiating with respect to x, we get:

$\frac{dy}{dx}$ = 4x3-18x2+26x-10

$(\frac{dy}{dx}) \bigg]_{(1,3)}$=4-18+26-10=2

Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as:

y-3=2(x-1)

=y-3=2x-2

=y=2x+1

The slope of the normal at (1, 3) is $\frac{-1}{\text{slop of the tangent at}\,(1,3)}$=-$\frac{1}{2.}$

Therefore, the equation of the normal at (1, 3) is given as:

y-3=-$\frac{1}{2.}$(x-1)

=2y-6=-x+1

x+2y-7=0

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(iii) The equation of the curve is y = x3 .

On differentiating with respect to x, we get:

$\frac{dy}{dx}$=3x2

$(\frac{dy}{dx}) \bigg]_{(1,1)}$=3(1)2=3

Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as:

y-1=3(x-1)

⇒ y=3x-2

The slope of the normal at (1, 1) is $\frac{-1}{\text{slope of the tanget at}\,(1,1)}$=$\frac{-1}{3.}$

Therefore, the equation of the normal at (1, 1) is given as:

y-1=$\frac{-1}{3.}$(x-1)

⇒3y-3=-x+1

⇒x+3y-4=0

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(iv) The equation of the curve is y = x2 .

On differentiating with respect to x, we get:

$\frac{dy}{dx}$=2x

$(\frac{dy}{dx}) \bigg]_{(0,0)}$=0

Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as: y − 0 = 0 (x − 0)

⇒ y = 0

The slope of the normal at (0, 0) is $\frac{-1}{\text{slop of the tangent at}\,(0,0)}=\frac{-1}{0}$, which is not defined.
Therefore, the equation of the normal at (x0, y0) = (0, 0) is given by
x=x0=0

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(v) The equation of the curve is x = cos t, y = sin t.

x=cost and y=sint

$\therefore\frac{dx}{dt}$=-sint,$\frac{dy}{dt}$=cost

=$\therefore\frac{dy}{dx}$=$\frac{(\frac{dy}{dt} )}{ (\frac{dx}{dt} )}$=$\frac{cost}{-sint}$=-cot t

$(\frac{dy}{dt}) \bigg]_{t=\frac{π}{4}}$=-cot t=-1

∴The slope of the tangent at t=$\frac{π}{4}$ is-1.

When t=$\frac{π}{4}$,x=$\frac{1}{√2}$and y=$\frac{1}{√2}$

Thus, the equation of the tangent to the given curve at t=$\frac{π}{4}$ i.e., at [($\frac{1}{√2}$,$\frac{1}{√2}$)] is

y-$\frac{1}{√2}$ =-1(x-$\frac{1}{√2}$).

x+y-$\frac{1}{√2}$-$\frac{1}{√2}$=0

x+y-√2=0

The slope of the normal at t=$\frac{π}{4}$ is $\frac{-1}{\text{slop of the tangent at} t= \frac{π}{4}}$ =1.

Therefore, the equation of the normal to the given curve at t=$\frac{π}{4}$ i.e., at[(\frac{1}{√2}$$\frac{1}{√2})] is

y-$\frac{1}{√2}$=1(x-$\frac{1}{√2}$).

⇒x=y