Derivatives: Find Q and its tangent

[zebo]

Homework Statement

Disclaimer: English is not my first language, so i apologize for any wrong math-terms.

We look at the function f(x) = x^3. On the graph for f we have a point, P(a,a^3), where a =/= 0. The tangent to f through P cuts through f in another point, Q. Find Q and show, that the tangent to f through Q has a slope which is 4 times as steep as the tangent to f through P.

Homework Equations

The Attempt at a Solution

Tangent through P cuts the function f in Q, which means in Q tangentP=x^3.

3a^2(x-a)+a^3=x^3 <=>
3a^2=(x^3-a^3)/(x-a) <=>
3a^2=x^2-a^2 <=>
4a^2=x^2
2a=x

Which means Q is located in (2a,(2a)^3)<=> Q(2a,8a^3)

But this means that the tangent through Q has a slope which is twice as steep as it should be, since:

tangentQ=8a^3+24a^2(x-a)

I am not sure how i should solve this problem, and i hope you are able to understand my sloppy english/and math.

 

[vela]

Homework Statement

Disclaimer: English is not my first language, so i apologize for any wrong math-terms.

We look at the function f(x) = x^3. On the graph for f we have a point, P(a,a^3), where a =/= 0. The tangent to f through P cuts through f in another point, Q. Find Q and show, that the tangent to f through Q has a slope which is 4 times as steep as the tangent to f through P.

Homework Equations

The Attempt at a Solution

Tangent through P cuts the function f in Q, which means in Q tangentP=x^3.

3a^2(x-a)+a^3=x^3 <=>
3a^2=(x^3-a^3)/(x-a) <=>

Looks fine up to here. However, you apparently think $\frac{x^3-a^3}{x-a} = x^2-a^2$. Try it with x=2 and a=1, for instance. You get 7 = 3. That doesn't work, does it?

3a^2=x^2-a^2 <=>
4a^2=x^2
2a=x

Which means Q is located in (2a,(2a)^3)<=> Q(2a,8a^3)

But this means that the tangent through Q has a slope which is twice as steep as it should be, since:

tangentQ=8a^3+24a^2(x-a)

I am not sure how i should solve this problem, and i hope you are able to understand my sloppy english/and math.

 

[BvU]

Hello Zebo, welcome to PF !

Your english is excellent, no problem reading and understanding at all. Now for the math :
If I make a sketch of f(x) = x^3 and draw a tangent at x=1 (so your a = 1), then I see that that tangent intersects the curve of f(x) at a negative value of x, and not at x= 2. Conclusion: there must be something wrong with your solving $3a^2 (x-a)+a^3 =x^3$. Do you agree ?

[edit]vela was faster but (fortunately )the replies match

 

[zebo]

Thank you for the fast response. Oh yeah i see, I am sorry about that i am snotty and heavy-headed.

My biggest problem right now is i have no clue which direction i need to take to solve this. I want to find Q, without using any calculators, and the chapter which this problem is a part of does not explain anything resembling this. Am i on the right path or is there a much simpler way?

Btw thanks again for the fast responses and thank you for welcoming me to the forum :)

 

[BvU]

Well, that is nice of you. So I'll throw in a hint: there is some information in the problem statement that you haven't used yet and that will give you a clue what the solution of this cubic equation might be ..

 

[vela]

My biggest problem right now is i have no clue which direction i need to take to solve this. I want to find Q, without using any calculators, and the chapter which this problem is a part of does not explain anything resembling this. Am i on the right path or is there a much simpler way?

You've taken a reasonable approach to the problem. You just need to get the algebra right.

As far as calculators go, we're not saying you need to use a calculator to solve the problem, but when you make mistakes, using a calculator to check and test your work is a good way to figure out where you went wrong. Also, looking at a specific example like a=1, can furnish you with insights on how to solve the general case. Solving a simpler but related problem can help you eventually solve the more general problem.

 

[zebo]

I've tried again:

3a^2=(x^3-a^3)/(x-a) <=> 3a^2 = x^2+a^2+ax <=> x^2-ax-2a^2 = 0

d = b^2 -4ac <=> d=9a^2

x = (a+3a)/2a v (a-3a)/2a <=> x = 2 v x = -1

So one of the x is where P is and the other is where Q is ? or have i gone down a wrong path?

I apologize if i am fumbling about with this, it is pretty late where i live

edit: If Q is to be found in (2,f(2)) the tangent would have a slope 4 times as steep as the tangent in P, and then P would be found in (-1,f(-1))?

 

[vela]

I've tried again:

3a^2=(x^3-a^3)/(x-a) <=> 3a^2 = x^2+a^2+ax <=> x^2-ax-2a^2 = 0

d = b^2 -4ac <=> d=9a^2

x = (a+3a)/2a v (a-3a)/2a <=> x = 2 v x = -1

You made a sign error somewhere. You should end up with $x = \frac{-a \pm |3a|}{2}.$ The $a$ that appears in the bottom of the quadratic formula is not the same as the $a$ you're using.

So one of the x is where P is and the other is where Q is ? or have i gone down a wrong path?

I apologize if i am fumbling about with this, it is pretty late where i live

edit: If Q is to be found in (2,f(2)) the tangent would have a slope 4 times as steep as the tangent in P, and then P would be found in (-1,f(-1))?

Remember how you defined your variables. You have $x$ as the x-coordinate of Q while $a$ is the x-coordinate of P.

 

[zebo]

You made a sign error somewhere. You should end up with $x = \frac{-a \pm |3a|}{2}.$ The $a$ that appears in the bottom of the quadratic formula is not the same as the $a$ you're using.Remember how you defined your variables. You have $x$ as the x-coordinate of Q while $a$ is the x-coordinate of P.

Thanks!

So as of now i have the following solution:

f(x) = x^3, P(a,a^3), a=/=0, f'(x)=3x^2

TangentP = a^3+3a^2(x-a)

In the point Q, TangentP = f(x)

a^3+3a^2(x-a) = x^3 <=> x^2+ax-2a^2 = 0

x = a v x = -2a

The tangent equals f(x) in P and Q, and P is located at (a,a^3). This means Q must be located at x = -2a

Q is located at (-2a,(-2a)^3) = Q(-2a,-8a^3)

The tangent in Q is the following:

TangentQ = -8a^3 + 3(-2a)^2(x-a) = -8a^3 + 12a^2(x-a)

The slope of tangentP was 3 and the slope of tangentQ is 12.

I conclude that Q is located at (-2a,-8a^3). The tangent to the point Q on f(x) has a slope that is 12/3 = 4 times as steep as the slope of tangentP.

 

[BvU]

Yes. The 4 is mentioned in the problem statement, so in fact you can get by just solving 3x^2 = 4 * 3 a^2 .

Old dutch saying: laziness makes inventive...

 

[zebo]

Yes. The 4 is mentioned in the problem statement, so in fact you can get by just solving 3x^2 = 4 * 3 a^2 .

Old dutch saying: laziness makes inventive...

Im not sure my math professor would approve, i did something similar before and he did not like it.

 

[BvU]

Sure, but once you know the answer is x = -2a you can check that it also satisfies the cubic equation (the other two solutions are x = a) and you don't have to tell explicitly how you found these factors...

from the picture it's evident

 

[zebo]

Thank you BvU and vela for your help :)