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 = x^{3 .}

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