A curve has equation y=2x^(3/2) +1. Find the point on the curve at which the gradient is 6, and then find the equation of the tangent to the curve at this point.

Question

mathmale · Answer

Quite an interesting problem!
You could do this problem in two different ways (at least):  (1) by finding the gradient or (2) by finding the derivative of the function in the usual way.
Start by explaining how to find the gradient of a function of two variables such as f(x,y).

anonymous · Answer

@mathmale Sorry for the late reply, if I find the derivative, which is f'(x) = 3x^(1/2), how would I then implement this in order to find the X and Y coordinates?

mathmale · Answer

Looks like we have to deal with the gradient in either case.
We're asked to determine where the gradient is 6, right?
My recollection of the gradient is that it is a VECTOR quantity, not a scalar one like 6.

Mind going back to ascertain that you've copied down the problem exactly as it was presented?

anonymous · Answer

Yes it is the exact wording of the problem. It is the second part of the problem however. The first part asked us to "calculate the gradient of the curve at: x=4 and x=16."

The weird thing is that the gradient of the curve at X=4 is 6, like the second part mentioned. So I do not know if I'm making this problem more difficult than it is and whether I just plug in f(4) = 2x^(3/2) +1, to find the Y-coordinate. (which would make it (4,17)

mathmale · Answer

"Yes it is the exact wording of the problem. It is the second part of the problem however. The first part asked us to "calculate the gradient of the curve at: x=4 and x=16."  Still, if you did that ("calc the grad at x=4, y-16), the gradient would be a VECTOR quantity.

So, how can we say that the gradient is 6, when 6 is a scalar, and the gradient is a vector, unless, of course, we take the MAGNITUDE  of the vector / gradient?

anonymous · Answer

I could attach a picture of the problem itself, as it is on my paper, maybe that could help..

anonymous · Answer

We are only into the calculus topic for about a month now, so haven't covered Vectors yet. Just the basics, like differentiating.

anonymous · Answer

@mathmale This is the questions as it appeared on my paper. If you could also confirm that my part 7a and b are correct, that would be brilliant. :) I got for a) 6 and for b) 12.

anonymous · Answer

@mathstudent55 Can you help me with problem 7 on the attached picture above? :)

mathmale · Answer

WD:  I've looked at your picture, especially at Problem #7.
A lot depends upon which course you're in.  Are you in a beginning calculus course, or are you in a "multivariable calculus" course in which functions are functions of 2, 3 or more variables?

some texts use "gradient" as an alternative word for slope.  I suspect that that is what you're learning.  My insistence on "gradient" being a vector quantity stems from a Multivariable calculus" point of view.

mathstudent55 · Answer

$\large y = x^{\frac{3}{2}} + 1$

$\large y' = \dfrac{3}{2}x^{\frac{1}{2}}$

$y' = \dfrac{3 \sqrt{x}} {2}$

anonymous · Answer

@mathmale I'm taking the IB HL Math course if that means anything to you and gradient is an alternative for slope in this case.

mathmale · Answer

what mathstudent55 has done here is to find the SLOPE of the given function's graph by finding the derivative.  As I said before, "gradient" is sometimes used in place of the word "slope."  mathstudent55's approach is just fine.

mathstudent55 · Answer

At x = 4, $f'(x) = f'(4) = \dfrac{3 \sqrt{4}}{2} = 3$

At x = 16, $f'(x) = f'(16) = \dfrac{3 \sqrt{16}}{2} = 6$

anonymous · Answer

@mathstudent55 It was 2x^3/2, though.

mathmale · Answer

I think you're both off on the right track and will let you work together.  I'll be in the background in case you have questions for me (but won't be responding immediately).

anonymous · Answer

@mathmale Alright, brilliant, thanks :)

anonymous · Answer

@mathstudent55 The curve is y=2x^3/2 +1. So that makes it 3x^1/2 when differentiated.

mathstudent55 · Answer

The power rule of differentiation is:

$f(x) = x^n$; $f'(x) = nx^{n - 1} $

mathstudent55 · Answer

Also, the derivative of a sum is the sum of the derivatives.
Finally, the derivative of a constant is zero.

mathstudent55 · Answer

Putting all those rules together, you get:

$\large f(x) = x^{\frac{3}{2}} + 1$

The derivative is:

$ f'(x) = \dfrac{d}{dx} ( x^{\frac{3}{2}} + 1)$

The derivative of a sum is the sum of the derivatives:

$ f'(x) = \dfrac{d}{dx} ( x^{\frac{3}{2}}) + \dfrac{d}{dx}(1)$

We use the power rule for the left derivative above.
The right derivative is the derivative of a constant which is zero.

$ \large f'(x) = \dfrac{3}{2} ( x^{\frac{1}{2}}) + 0$

$ \large f'(x) = \dfrac{3}{2} ( x^{\frac{1}{2}})$

$ \large f'(x) = \dfrac{3 \sqrt{x}}{2} $

anonymous · Answer

I understand that the derivative of a constant is zero, but what I was trying to say is that on the question itself, it says that the curve is y=2x^3/2 +1, not y=x^3/2+1.

anonymous · Answer

So when we differentiate:

$$f(x)=2x ^{3/2}+1$$

We get:

$$f'(x)=3\sqrt{x}$$

mathstudent55 · Answer

Sorry. I see now. There was a factor of 2 that i missed.
You are correct. The derivative is 

$y' = 3\sqrt{x} $

anonymous · Answer

No problem, I think I got it now :). Thanks for all the help, really appreciate. Here's a best response thing and a new fan :D

mathstudent55 · Answer

You're welcome.