Question

calc help!

Answer 1

@AccessDenied

Answer 2

I think your first step will be finding \( \dfrac{d^2 y}{dx^2} \). After you have that, you should be able to plug in the values of x and y.

Do you know how to use implicit differentiation on this equation?
\( 3x^2 + y^2 = 7 \)

Answer 3

no...

Answer 4

Ah. Implicit differentiation is all about the chain rule. When you have some function of y, you are just differentiating with respect to y and then multiplying by dy/dx.

Like this: \( \dfrac{d}{dx} f(y) = \dfrac{df(y)}{\color{blue}{dy}} \times \dfrac{\color{blue}{dy}}{dx} \)

Answer 5

ohh that doestn look too bad

Answer 6

In this problem, we begin by taking the derivative of both sides with respect to x:
\( \dfrac{d}{dx} \left( 3x^2 + y^2 \right) = \dfrac{d}{dx} \left( 7 \right) \)
The derivative "distributes" through addition. So we will be taking the derivative term-by-term:
\( \dfrac{d}{dx} \left( 3x^2 \right) + \dfrac{d}{dx} \left( y^2 \right) = \dfrac{d}{dx} \left( 7 \right) \)
The first derivative on the left is just power rule. The second is a function of y, so we do use that chain rule idea above. The last is just a constant, so it's derivative is 0.
\( 6x + 2y \times \dfrac{dy}{dx} = 0 \)

Answer 7

Anything there not entirely clear or should be explained a little more? :)

Answer 8

so the answer is just 0? theres nothing else to do after that?

Answer 9

that doesnt look hard at all hahaha

Answer 10

No, the problem is not quite done yet! Right now this equation just relates x, y, and dy/dx. If you plugged in x=1 and y=2, you could solve for dy/dx but that's not what they wanted. :P

What we have to do next is, basically, take another derivative of both sides.

Answer 11

ohh now i see how this is confusing hahaha

Answer 12

\( \dfrac{d}{dx} \left( 6x + 2y \dfrac{dy}{dx} \right) = \dfrac{d}{dx} \left( 0 \right) \)
Now it gets just a little more fun because we have that dy/dx. But not all is lost! :P

Answer 13

Again we distribute the d/dx through. d/dx(0) = 0, just another constant.
\( \dfrac{d}{dx} \left( 6x \right) + \dfrac{d}{dx} \left( 2y \dfrac{dy}{dx} \right) = 0 \)

Answer 14

Here's where it gets just a little tricky!
The first derivative there is just another power rule, not too bad. \( \dfrac{d}{dx} 6x = 6 \)
The second item there is a product of two functions: \( 2y\) and \( \dfrac{dy}{dx} \)!

We have to apply product rule now:
\( \color{red}{\dfrac{d}{dx}} \left( 2y \dfrac{dy}{dx} \right) = \color{red}{\dfrac{d}{dx}}\color{#aa0000}{\left( 2y \right)} \dfrac{dy}{dx} + 2y \color{red}{\dfrac{d}{dx}} \color{#aa0000}{\left( \dfrac{dy}{dx} \right)} \)
Tried to highlight that step so it makes visual sense what is going on.

Answer 15

your explanation is honestly the best

Answer 16

The derivative of a derivative becomes that 2nd derivative item. \( \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right) = \dfrac{d^2y}{dx^2} \)
So putting it all together, our final equation becomes:

\( 6 + 2 \cdot \dfrac{dy}{dx} \cdot \dfrac{dy}{dx} + 2y \cdot \dfrac{d^2y}{dx^2} = 0\)

Answer 17

So our last step is just plugging in the x = 1, y = 2. But you might wonder, what about those dy/dx?

You'll have to plug it in here to get the dy/dx value: \( 6x + 2y \times \dfrac{dy}{dx} = 0 \)
\( 6\cdot (1) + 2 \cdot (2) \times \dfrac{dy}{dx} = 0 \) We solve for dy/dx here
\( 6 + 4 \dfrac{dy}{dx} = 0 \)
\( \dfrac{dy}{dx} = -\dfrac{2}{3} \)

And plug dy/dx = -2/3, x=1, and y=2 into that d^2y/dx^2 equation; solve for d^2y/dx^2 !

Answer 18

Oh woops, I'm being silly. \( \dfrac{dy}{dx} = -\dfrac{3}{2} \)
You subtract 6 and then divide by 4. :P

Answer 19

\( 6 + 2 \cdot \dfrac{dy}{dx} \cdot \dfrac{dy}{dx} + 2y \cdot \dfrac{d^2y}{dx^2} = 0 \)
x=1, y=2, dy/dx = -3/2
\( 6 + 2 \cdot \left(-\dfrac{3}{2} \right) \left( - \dfrac{3}{2} \right) + 2 (2) \dfrac{d^2y}{dx^2} = 0 \)
All that is left is to solve for \( \dfrac{d^2y}{dx^2} \) and we get the final answer at last!

Answer 20

i know what x and y is to plug in, but what is d, im confused

Answer 21

dy/dx is the derivative of y with respect to x, or \( \dfrac{dy}{dx} \). The d is, in math terms, an "operator" that says, take the derivative of whatever comes after it.

Answer 22

but youre saying to solve for d^2y/dx^2

Answer 23

i really dont know

Answer 24

Yup. We solve for that whole item there.
\( 6 + 2 \cdot \left(-\dfrac{3}{2} \right) \left( - \dfrac{3}{2} \right) + 2 (2) \color{blue}{\dfrac{d^2y}{dx^2}} = 0 \)
Just as if it were a single variable, like A.
\( 6 + 2 \cdot \left(-\dfrac{3}{2} \right) \left( - \dfrac{3}{2} \right) + 2 (2) \color{blue}A = 0 \)

We'd subtract off 6 and 9/2 (2*-3/2 *-3/2), and then divide by 2*2=4 to solve for that value.

Answer 25

i got 22

Answer 26

I will write out the steps:
\( 6 + 2 \cdot \left( -\dfrac{3}{2} \right) \left( -\dfrac{3}{2} \right) + 2(2) \dfrac{d^2 y}{dx^2} = 0 \)
Simplify a bit:
\( 6 + 4.5 + 4 \dfrac{d^2 y }{dx^2} = 0 \)
\( 10.5 + 4 \dfrac{d^2 y}{dx^2} = 0 \)
Subtract both sides by 10.5
\( 4 \dfrac{d^2 y}{dx^2} = -10.5 \)
\( \dfrac{d^2y}{dx^2} = \dfrac{-10.5}{4} \)
You could use a calculator here since the answer must be two decimal places. I get about -2.63.

Answer 27

yes i got th same, thank you (:

Answer 28

Glad to help out! :)