1) Let x^3 + y^3 = 28. Find y"(x) at the point (3, 1). y"(3)=
The correct answer is: -2*3*28 = -168
3x^2 + 3y^2(dx/dy) = 0
6x + 6y(dx/dy) = 0
dx/dy = -6(3)/6(1) = -3
That's as far as i get... i don't even know if that's started right...
You have
x^3 + + y^3 = 28
3x^2 + 3y^2(dy/dx) = 0 (that should be dy/dx) so
dy/dx = -x^2 / y^2
Now evaluate y=1 x=3 so y'(3)=-9
Now we want y"
d2y/dx2 = [y^2(-2x) - -x^2*2y*y']/y^4
Now just evaluate
y"(3)= [-6 - 9*2*9]/1 = -168
Verify the algebra is correct.
good work.
Thnx, but in hindsight I'm wondering how much the students learn when I do the entire problem. I really need to practice giving just the set-up and letting them complete it. Gotta' admit, these problems are fun, now. I can recall struggling with this stuff just like some of them are, so I keep telling people to persevere until you 'breakthrough'.
Well, breaking through is definitely the key! It's like trying to open a jar of pickles. You just keep twisting and twisting, and eventually, POP! The lid comes off and all that deliciousness is yours. Just like with math, you gotta keep twisting those gears in your brain until you find the right solution. And hey, if all else fails, you can always invite some friends over for a jar of pickles and a good laugh. Who knew math could be so tasty?
It seems like you have made some progress in finding the second derivative of y with respect to x. Here, I will provide a step-by-step explanation to help you complete the calculation.
Given the equation x^3 + y^3 = 28, we need to find y"(x) at the point (3, 1).
To find the first derivative, differentiate both sides of the equation with respect to x:
d/dx(x^3 + y^3) = d/dx(28)
Using the chain rule, we have:
3x^2 + 3y^2 * dy/dx = 0
To isolate dy/dx, we can rearrange the equation:
dy/dx = -3x^2 / (3y^2)
Now, we can substitute the coordinates of the point (3, 1) into the equation to find dy/dx:
dy/dx = -3(3)^2 / (3(1)^2)
= -27/3
= -9
So, the slope of the tangent line at the point (3, 1) is -9.
To find the second derivative, we need to differentiate the expression dy/dx = -3x^2 / (3y^2) with respect to x. Applying the quotient rule, we have:
d^2y/dx^2 = [y^2 * (-2x) - (-3x^2) * 2y * dy/dx] / y^4
= [-2xy^2 + 6x^2y * dy/dx] / y^4
Now, substitute the coordinates of the point (3, 1) and the value of dy/dx into the equation:
y"(3) = [-2(3)(1)^2 + 6(3^2)(1)(-9)] / (1^4)
= [-6 + 162(-9)] / 1
= (-6 - 1458) / 1
= -1464 / 1
= -1464
Therefore, the value of y"(3) at the point (3, 1) is -1464.
It seems like there was a mistake in the previous answer you mentioned (-168). Please double-check the calculations to verify the accuracy, and let me know if you have any further questions.
To find y"(x) at the point (3, 1) for the equation x^3 + y^3 = 28, we can follow these steps:
1. Start with the equation x^3 + y^3 = 28.
2. Differentiate both sides of the equation with respect to x using the chain rule. This gives us:
3x^2 + 3y^2(dy/dx) = 0
Simplify the equation to:
x^2 + y^2(dy/dx) = 0
3. Solve for dy/dx to find the first derivative of y with respect to x. Rearranging the equation, we get:
dy/dx = -x^2 / y^2
4. Evaluate dy/dx at the point (3, 1). Substitute x = 3 and y = 1 into the equation:
dy/dx = -(3^2) / (1^2)
= -9
Therefore, y'(3) = -9.
5. To find the second derivative, differentiate dy/dx with respect to x. Using the quotient rule, the equation becomes:
y''(x) = [y^2(-2x) - -x^2 * 2y * dy/dx] / y^4
Simplifying further, we have:
y''(x) = [-2xy^2 + 2x^2y(dy/dx)] / y^4
6. Substitute the values x = 3, y = 1, and dy/dx = -9 into the equation:
y''(3) = [-2(3)(1^2) + 2(3^2)(1)(-9)] / (1^4)
= [-6 - 162] / 1
= -168
Therefore, y''(3) = -168.
So, the second derivative of y with respect to x at the point (3, 1) is -168.