Please Help With This Question!!! (1 Viewer)

[star*eyed]

Find the equation of the normal (n) to the curve
Y= x <SUP>4</SUP> + 4x <SUP>3/2</SUP> at the point A (1,5)


How do I do this question??? im really really stuck!

 

[SonyHK]

Find the equation of the normal (n) to the curve
Y= x <SUP>4</SUP> + 4x <SUP>3/2</SUP> at the point A (1,5)


How do I do this question??? im really really stuck!

Y'=4x³+6x^(1\2)
Equation of the normal to the curve is (y-5)\(x-1)=-1\[4(1)³+6(1)^(1\2)]=-1\10

10(y-5)=-(x-1)
10y-50=-x+1
.: x+10y-51=0 is the required equation, if there are no errors.

 

[acullen]

Firstly we differentiate the function with respect to x to find the gradient at any point:
y=x⁴+4x^3/2
dy/dx = 4x³+6x^1/2

Now to obtain the gradient of the point A (1,5), substitue in the value for x.
dy/dx=4*1³+6*1^1/2
=4+6
=10

Now the normal m_n of the gradient of a tangent m_t is equal to:
m_n = -m_t^-1
= -10^-1
=-0.1

Now we have our gradient for our normal to the curve, we use the linear form of a line and substitue in points to obtain the constant:
y=mx+b

where:
m = -0.1
y = 5
x = 1

5=-0.1*1+b
b=5.1
.'. y=-0.1x+5.1

or in the general form, just multiply every term by -10 and solve for x+ny+d=0
x+10y-51=0

 

[acullen]

Finding the result the way the person above me did is pretty much the same method I used; however, use the formula y-y₁=m(x-x₁) in lieu of y=mx+b that I used. Either way you obtain the correct answer, so it's up to you which formula you use to solve these problems.

 

[star*eyed]

Thanx Heaps For All The Help!

 

[YBK]

Finding the result the way the person above me did is pretty much the same method I used; however, use the formula y-y₁=m(x-x₁) in lieu of y=mx+b that I used. Either way you obtain the correct answer, so it's up to you which formula you use to solve these problems.

yup, that's true.

but my teacher gets really pissy when I use y = mx + b for that type of question. she says it takes a lot longer... which is sorta true.