Question regarding finding the tangent line to a point on a curve

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September 30th 2011, 09:59 AM
iAmKrizzle

Question regarding finding the tangent line to a point on a curve

So, I have been working problems finding the derivative of polynomial equations. I am having some problems, however. The problem is as follows:

Find the line tangent to the curve of Y = 4th root of X, on the point (1, 1).

This is my working of the problem...
y= 4th root of x
y' = x ^ 1/4 = 1/4 x ^ -3/4 = m

therefore, the line tangent to this should be Y - Y1 = m(x - x1) so:
y - 1 = 1/4 x ^ -3/4(x - 1)
y - 1 = 1/4 x ^ 1/4 - 1/4x ^ -3/4
y = 1/4 x ^ 1/4 - 1/4x ^ -3/4 + 1
y = 1/4x + 1 <<<<<<

So by my calculations the line equation should be y = 1/4x + 1. However the book has the answer as y = 1/4x + 3/4. Where have I gone wrong? I don't see it.
September 30th 2011, 10:11 AM
Plato

Re: Question regarding finding the tangent line to a point on a curve

Quote:

Originally Posted by iAmKrizzle

Find the line tangent to the curve of Y = 4th root of X, on the point (1, 1).

I cannot read what you posted.
If $y=x^{\frac{1}{4}}$ then $y'=\frac{1}{4}x^{\frac{-3}{4}}$ .

So at (1,1) the slope is $\frac{1}{4}$ .

Tangent line is $(y-1)=\frac{1}{4}(x-1)$
September 30th 2011, 11:05 AM
HallsofIvy

Re: Question regarding finding the tangent line to a point on a curve

Quote:

Originally Posted by iAmKrizzle

So, I have been working problems finding the derivative of polynomial equations. I am having some problems, however. The problem is as follows:

Find the line tangent to the curve of Y = 4th root of X, on the point (1, 1).

This is my working of the problem...
y= 4th root of x

Okay, y= x^1/4)

Quote:

y' = x ^ 1/4

Oh dear, no! If y= x^(1/4) then y' is NOT the same thing!
I know that's not what you meant to say, but is is what you actually said. Be careful what you write. Mathematics is very precise.
What you meant to say was

Quote:

y'= 1/4 x ^ -3/4 = m

therefore, the line tangent to this should be Y - Y1 = m(x - x1) so:
y - 1 = 1/4 x ^ -3/4(x - 1)

No, the slope or derivative at x= 1 is (1/4)1^(-3/4)= 1/4. It is a constant, not a function of x.

Quote:

y - 1 = 1/4 x ^ 1/4 - 1/4x ^ -3/4
y = 1/4 x ^ 1/4 - 1/4x ^ -3/4 + 1
y = 1/4x + 1 <<<<<<

So by my calculations the line equation should be y = 1/4x + 1. However the book has the answer as y = 1/4x + 3/4. Where have I gone wrong? I don't see it.

y- 1= (1/4)(x- 1)= (1/4)- 1/4. Adding 1 to both sides, y= (1/4)x+ 1- 1/4= (1/4)x+ 3/4.
September 30th 2011, 01:13 PM
iAmKrizzle

Re: Question regarding finding the tangent line to a point on a curve

HallsofIvy and Plato,

Thanks your for your responses I see now where I went astray. With your help I've come to the correct answer also. Thank you again.