Question

Calculus. The tangent to the curve y=ax^4 at the point (3,b) has a gradient of 2. Find the values of a and b. Please show working. :)

Answer 1

(3, b) lies on the curve \(y=ax^4\)
so
\[b=a\times3^4..................(1)\]
It's given the gradient of the tangent is 2 at (3, b), that is
\[\large (\frac{dy}{dx})_{(3, b)}=2\]
Find dy/dx , put x=3 and y=b, and equate it to 2
you'll get the second equation.
Two equations , two variables
voila!

Answer 2

Do let me know if you have any doubt:)

Answer 3

@Lachlan1996 Do you get this?

Answer 4

Umm somewhat, do excuse me, i am a little slow with this. we are only just into year 11 calculus. I do follow somewhat but im not sure what you mean by equating (dydx)(3,b)=2

Answer 5

Okay, we'll do it step by step
firstly Find dy/dx

Answer 6

thats the bit im not sure with

Answer 7

Do you know differentiation?

Answer 8

i am familiar with the basics, but we are still learning it

Answer 9

@Lachlan1996 I gotta bounce now, Sorry
My friend @apoorvk will help you on this

Answer 10

okie doke, mate, thanks for the help so far

Answer 11

Hmm... okay so do you want me start all over again? I can do that, not a problem..

Answer 12

Do whatever suits you mate, im happy with any help

Answer 13

Okay, I'll proceed a wee bit of recap.
So, did you get the part where we satisfy (3,b) in the equation of the curve \(y=ax^4\), since the point lies on the curve?

Answer 14

yes. The x coordinate 3 is substituted into the formula [y=ax^4] to get a*81

Answer 15

So,
b = 81a ....(i)

This is one equation that we have. Okay good.
Now now..

We need a general equation of the slope.gradient value for the curve. What do we do - shoul I differentiate \(y=ax^4\)??

Answer 16

*should

Answer 17

uhh \[y=ax^4\] = \[4ax^3\] doesn't it?

Answer 18

yeah, means:\[y' = \frac{d y}{dx}= 4ax^3\]

Now, dy/dx, that is the " \(4ax^2 \) " obtained, is the general expression for the value of slope/gradient for the curve \(ax^4\) at any particular 'x'.

Okay till here?

Answer 19

do you mean "ax^3

Answer 20

oh yeah, I meant \(4ax^3\). Typo sorry.

Answer 21

and so 4(a)(3)^3-1 = the gradient of this point

Answer 22

no sorry typo there too

Answer 23

and so 4(a)(3)^4-1

Answer 24

Umm. you mean \(4a(3)^{4-1}\) right? That's the slope at (3,b)

Answer 25

yes thats what i meant

Answer 26

and so 108a=2. thereby a=1/54

Answer 27

and so b = 1/54 *81

Answer 28

Hmm, and now, according to the question the value of the gradient at (3,b) is '2'.
So,
\(4a3^3 = 108a = 2\)
And yes, so a = 1/54

Answer 29

Thankyou very much for your help :D

Answer 30

And we had an equation in the beginning remember?
b= 81a
so, b = 81/54 = 3/2

That's it! You are very welcome, and you're a good learner!

Answer 31

your a top bloke mate :) cheers for the shunt i needed. Helped me out a lot. Big thanks to you!

Answer 32

hey man, not me, it was your own desire to learn. Cheerio ;)

Answer 33

Have a good night mate :) wish you well

Answer 34

Good night to you as well! Let us know if we can help you anymore! :)

Answer 35

Will do mate. Gotta knock the maths off, then chem and physics

Answer 36

Kick 'em hard! Good luck!

Answer 37

Will do. Big thanks to you, and give my thanks to ash

Answer 38

sure mate, it was his work on this after all ;)

Answer 39

The tangent to the curve y=ax^3 at the point (-1,b) is perpendicular to the line y=2x+3 Find the values of a and b. I would take this in the same fashion by finding the gradient first yes?