MHB Further question on My Fractions problem

[tmt1]

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]

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?

 

[tmt1]

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]

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:

$$x^2+(y-x)^3=9$$

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

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

Solving for $$\frac{dy}{dx}$$, we find:

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

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

$$1^2+(y-1)^3=9$$

$$y=3$$

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

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

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

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

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

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

View attachment 2098

I agree with your answer. (Yes)

 

[Deveno]

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:

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