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Further question on My Fractions problem

tmt

Active member
Jan 15, 2014
236
I have a separate question on the same problem from my prior post.

I need an equation for a tangent which has a slope of 5/6 and passes through (1,3)

y-3 = 5/6(x-1)

I simplify this to

y= 5/6x +13/6

However the answer given is y = 7/6(x) + 13/6

Where am I going wrong?

Yours,

Timothy
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Your line has the required slope, while the given answer has the wrong slope. It is most likely a typo somewhere, either in the statement of the problem or the given answer. Can you post the original problem in its entirety?
 

tmt

Active member
Jan 15, 2014
236
PROBLEM 11 : Find an equation of the line tangent to the graph of x2 + (y-x)3 = 9 at x=1 . '


This is the end of the answer:

Thus, the slope of the line tangent to the graph at (1, 3) is

$ m = y' = \displaystyle{ 3 (3-1)^2 - 2(1) \over 3 (3-1)^2 } = \displaystyle{ 10 \over 12 } = \displaystyle{ 5 \over 6 } $ ,

and the equation of the tangent line is

y - ( 3 ) = (5/6) ( x - ( 1 ) ) ,

or

y = (7/6) x + (13/6) .

I suspect it is a typo in the answer. Here is the link for the full answer.

https://www.math.ucdavis.edu/~kouba...soldirectory/ImplicitDiffSol.html#SOLUTION 11
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Okay, I see now...I assumed the slope was given as 5/6. Let's take a look at the problem. We are given the curve:

\(\displaystyle x^2+(y-x)^3=9\)

So, implicitly differentiating with respect to $x$, we find:

\(\displaystyle 2x+3(y-x)^2\left(\frac{dy}{dx}-1 \right)=0\)

Solving for \(\displaystyle \frac{dy}{dx}\), we find:

\(\displaystyle \frac{dy}{dx}=1-\frac{2x}{3(y-x)^2}\)

Now, when $x=1$, we find from the original curve:

\(\displaystyle 1^2+(y-1)^3=9\)

\(\displaystyle y=3\)

And so we find the slope at the given point is:

\(\displaystyle \left.\frac{dy}{dx} \right|_{(x,y)=(1,3)}=1-\frac{2(1)}{3(3-1)^2}=1-\frac{2}{12}=\frac{5}{6}\)

Hence, using the point-slope formula, we obtain the tangent line:

\(\displaystyle y-3=\frac{5}{6}(x-1)\)

\(\displaystyle y=\frac{5}{6}x+\frac{13}{6}\)

Here is a plot of the curve and the tangent line:

tmt.jpg

I agree with your answer. (Yes)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
PROBLEM 11 : Find an equation of the line tangent to the graph of x2 + (y-x)3 = 9 at x=1 . '


This is the end of the answer:

Thus, the slope of the line tangent to the graph at (1, 3) is

$ m = y' = \displaystyle{ 3 (3-1)^2 - 2(1) \over 3 (3-1)^2 } = \displaystyle{ 10 \over 12 } = \displaystyle{ 5 \over 6 } $ ,

and the equation of the tangent line is

y - ( 3 ) = (5/6) ( x - ( 1 ) ) ,

or

y = (7/6) x + (13/6) .

I suspect it is a typo in the answer. Here is the link for the full answer.

https://www.math.ucdavis.edu/~kouba...soldirectory/ImplicitDiffSol.html#SOLUTION 11
I concur that both you and MarkFL are correct, the link you provided has a typo in the very last line (it is correct until that), and the correct tangent line has the equation:

$y = \dfrac{5}{6}x + \dfrac{13}{6}$