Approximating slope of the derivative at a point

In summary, the slope of the line tangent to the curve y^3 + x^2y^2 - 3x^3 = 9 at (1.5,2) is approximately 0.39. The derivative was solved using implicit differentiation, but a small error was made in dropping an exponent on the y in the denominator. The correct derivative is (9x^2-2xy^2)/(3y^2+2x^2y), which when plugged in results in a slope of 0.39.
  • #1
demersal
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Homework Statement


The slope of the line tangent to the curve y^3 + x^2y^2 - 3x^3 = 9 at (1.5,2) is approximately


Homework Equations


Answer choices
A) 0.39
B) 3.2
C) -11.45
D) -2.29
E) -3.2


The Attempt at a Solution


Well, I solved for the derivative, but I'm not sure if it was to any avail seeing as when I plug the point back in I do not get an answer choice.

I got:

dy/dx = (-2xy^(2) + 9x^2) / (3y + 2x^(2)y)

Did I solve the derivative incorrectly? Or is plugging in the point not the correct course of action? Thank you again for all your help, I've been struggling with this problem a lot and any guidance/explanation would be great!
 
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  • #2
Hello,

It looks like you made a small error in your implicit differentiation-- go back and double check, see if you can spot it.

Hope this helps.
 
  • #3
Yep, I think you simply dropped an exponent on the y from the denominator. I got

[tex]\frac{dy}{dx}=\frac{9x^{2}-2xy^{2}}{3y^{2}+2x^{2}y}[/tex]

Plugging in the numbers, I get .393. Hope that helps.
 
Last edited:

1. What is the definition of the derivative at a point?

The derivative at a point is the slope of the tangent line to the curve at that point. It represents the instantaneous rate of change of a function at a specific point.

2. How is the slope of the derivative at a point approximated?

The slope of the derivative at a point can be approximated using the limit definition of the derivative, where the slope is calculated as the change in y-values over the change in x-values as the interval around the point becomes smaller and smaller.

3. Why is it important to approximate the slope of the derivative at a point?

Approximating the slope of the derivative at a point allows us to understand the behavior of a function at a specific point and make predictions about its future values. It also helps us analyze the rate of change of a function and identify any critical points.

4. What is the relationship between the slope of the derivative and the concavity of a function?

The slope of the derivative at a point can give us information about the concavity of a function. If the slope is positive, the function is increasing and the concavity is upward. If the slope is negative, the function is decreasing and the concavity is downward. A slope of zero indicates a horizontal tangent line and a possible point of inflection where the concavity changes.

5. Can the slope of the derivative at a point be negative and zero at the same time?

No, the slope of the derivative at a point cannot be negative and zero simultaneously. It can either be positive, negative, or zero, but not a combination of both. A slope of zero indicates a horizontal tangent line, while a negative slope indicates a downward direction of the curve.

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