Question

Find the equations of the tangent and the normal to the curve y= 5x^4 at the point (1,5)​

Answer 1

SOLUTION

TO DETERMINE

The the equations of the tangent and the normal to the below curve at the point (1,5)

$\displaystyle\sf{y = 5 {x}^{4} }$

EVALUATION

Here the given equation of the curve is

$\displaystyle\sf{y = 5 {x}^{4} }$

Differentiating both sides with respect to x we get

$\displaystyle\sf{ \frac{dy}{dx} = 2 0{x}^{3} }$

Now

$\displaystyle\sf{ \frac{dy}{dx} \bigg|_{(1 ,5)} = 2 0 }$

Hence the required equation of the tangent at the point (1,5) is

$\sf{(y - 5) = 20(x - 1)}$

$\sf{ \implies \: (y - 5) = 20x -20}$

$\sf{ \implies \: 20x - y = 15 }$

Now the equation of the line perpendicular to the above line is

$\sf{ 20y + x = c }$

Now the line passes through the point (1,5)

Thus we get

$\sf{ \implies \: 100 + 1= c }$

$\sf{ \implies \: c = 101 }$

Hence the required equation of the normal is

$\sf{ x + 20y = 101}$

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Answer 2

Step-by-step explanation:

SOLUTION

TO DETERMINE

The the equations of the tangent and the normal to the below curve at the point (1,5)

\displaystyle\sf{y = 5 {x}^{4} }y=5x4

EVALUATION

Here the given equation of the curve is

\displaystyle\sf{y = 5 {x}^{4} }y=5x4

Differentiating both sides with respect to x we get

\displaystyle\sf{ \frac{dy}{dx} = 2 0{x}^{3} }dxdy=20x3

Now

\displaystyle\sf{ \frac{dy}{dx} \bigg|_{(1 ,5)} = 2 0 }dxdy∣∣∣∣∣(1,5)=20

Hence the required equation of the tangent at the point (1,5) is

\sf{(y - 5) = 20(x - 1)}(y−5)=20(x−1)

\sf{ \implies \: (y - 5) = 20x -20}⟹(y−5)=20x−20

\sf{ \implies \: 20x - y = 15 }⟹20x−y=15

Now the equation of the line perpendicular to the above line is

\sf{ 20y + x = c }20y+x=c

Now the line passes through the point (1,5)

Thus we get

\sf{ \implies \: 100 + 1= c }⟹100+1=c

\sf{ \implies \: c = 101 }⟹c=101

Hence the required equation of the normal is

\sf{ x + 20y = 101}x+20y=101