Math Is Fun Forum

mathland

Member

Registered: 2021-03-25

Posts: 444

Tangent Line

Let T be the line tangent to the graph of y = x^3 at the point (1/2, 1/8).
At what other point Q on the graph
of y = x^3 does the line T intersect the graph? What is the slope of
the tangent line at Q?

NOTE: I am not seeking the answer but the set up only. I will do the math work.

Thank you.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,640

Re: Tangent Line

work out dy/dx for the gradient of the tangent line and hence find the equation using y = mx + c.

This line will cross the curve again so work with a pair of simultaneous equations (tangent equation) and (curve equation).

The resulting equation in x will look complicated but you have one extra clue ... you know (1/2 , 1/8) lies on both so the expression must factorise with (x-1/2) or (2x-1) as a factor.  That leaves a quadratic implies two more solutions.  Wait a mo though.  If the line makes a tangent at that point then (2x-1) will be a double factor leaving only a linear equation left for the solution you want.

B.

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathland

Member

Registered: 2021-03-25

Posts: 444

Re: Tangent Line

work out dy/dx for the gradient of the tangent line and hence find the equation using y = mx + c.

This line will cross the curve again so work with a pair of simultaneous equations (tangent equation) and (curve equation).

The resulting equation in x will look complicated but you have one extra clue ... you know (1/2 , 1/8) lies on both so the expression must factorise with (x-1/2) or (2x-1) as a factor.  That leaves a quadratic implies two more solutions.  Wait a mo though.  If the line makes a tangent at that point then (2x-1) will be a double factor leaving only a linear equation left for the solution you want.

B.

This is one is a bit unclear.

Mathegocart

Member

Registered: 2012-04-29

Posts: 2,226

Re: Tangent Line

work out dy/dx for the gradient of the tangent line and hence find the equation using y = mx + c.

This line will cross the curve again so work with a pair of simultaneous equations (tangent equation) and (curve equation).

The resulting equation in x will look complicated but you have one extra clue ... you know (1/2 , 1/8) lies on both so the expression must factorise with (x-1/2) or (2x-1) as a factor.  That leaves a quadratic implies two more solutions.  Wait a mo though.  If the line makes a tangent at that point then (2x-1) will be a double factor leaving only a linear equation left for the solution you want.

B.

This is one is a bit unclear.

I hope this explanation of mine may help in clarifying Bob's explanation to you.

To find that specific point Q, we need to find the equation of the tangent line T(y = mx + b) at the point(1/2, 1/8).
So, let's find the derivative of the function y = x^3 so we can find the slope of said line.
y' = 3x^2(per the power rule.)
Plug in x = 1/2 and we get y' = 3(1/2)^2 = 3/4. So the slope, or "m," in the tangent line is 3/4.

Now we have y = (3/4)x + b. Plug in the point that the tangent line contains so you can find b, and with that equation, of the tangent line you want to find where it intersects with y = x^3 so you can determine point Q.

Set the ys equal and you will receive
The Tangent Line Equation = x^3.

As Bob mentions, you will receive a cubic equation that can be reduced down to a quadratic equation via dividing the cubic by (2x-1)(as you already know that x = 1/2 is a solution to the problem above.)

You will find the other solution to the quadratic, and then determining the slope of the tangent line at Q to y=x^3 at x = other solution is relatively easy.

Last edited by Mathegocart (2021-04-19 13:59:51)

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The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.

mathland

Member

Registered: 2021-03-25

Posts: 444

Re: Tangent Line

work out dy/dx for the gradient of the tangent line and hence find the equation using y = mx + c.

This line will cross the curve again so work with a pair of simultaneous equations (tangent equation) and (curve equation).

The resulting equation in x will look complicated but you have one extra clue ... you know (1/2 , 1/8) lies on both so the expression must factorise with (x-1/2) or (2x-1) as a factor.  That leaves a quadratic implies two more solutions.  Wait a mo though.  If the line makes a tangent at that point then (2x-1) will be a double factor leaving only a linear equation left for the solution you want.

B.

This is one is a bit unclear.

I hope this explanation of mine may help in clarifying Bob's explanation to you.

To find that specific point Q, we need to find the equation of the tangent line T(y = mx + b) at the point(1/2, 1/8).
So, let's find the derivative of the function y = x^3 so we can find the slope of said line.
y' = 3x^2(per the power rule.)
Plug in x = 1/2 and we get y' = 3(1/2)^2 = 3/4. So the slope, or "m," in the tangent line is 3/4.

Now we have y = (3/4)x + b. Plug in the point that the tangent line contains so you can find b, and with that equation, of the tangent line you want to find where it intersects with y = x^3 so you can determine point Q.

Set the ys equal and you will receive
The Tangent Line Equation = x^3.

As Bob mentions, you will receive a cubic equation that can be reduced down to a quadratic equation via dividing the cubic by (2x-1)(as you already know that x = 1/2 is a solution to the problem above.)

You will find the other solution to the quadratic, and then determining the slope of the tangent line at Q to y=x^3 at x = other solution is relatively easy.

I will play with one some more.