Civil Engineering (CE) Exam > Civil Engineering (CE) Questions > For the curve xy3- yx3= 6, the slope of the t...
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For the curve xy3 - yx3 = 6, the slope of the tangent line at the point (1, -1) is:
- a)-1
- b)1/2
- c)2
- d)1
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
For the curve xy3- yx3= 6, the slope of the tangent line at the point ...
Problem Analysis:
We are given the equation of a curve: xy^3 - yx^3 = 6. We need to find the slope of the tangent line at the point (1, -1).
Solution:
To find the slope of the tangent line at a given point on the curve, we need to take the derivative of the curve equation with respect to x and then substitute the x-coordinate of the given point into the derivative equation.
Let's differentiate the equation xy^3 - yx^3 = 6 with respect to x.
Differentiating the equation:
To differentiate the equation, we will use the product rule and the chain rule of differentiation.
The product rule states that if u(x) = f(x)g(x), then the derivative of u(x) with respect to x is given by:
u'(x) = f'(x)g(x) + f(x)g'(x)
The chain rule states that if y = f(u) and u = g(x), then the derivative of y with respect to x is given by:
dy/dx = dy/du * du/dx
Differentiating xy^3 - yx^3 = 6 with respect to x using the product rule, we get:
y^3 + 3xy^2 * dy/dx - 3x^2y - y^3 * dx/dx = 0
Simplifying the equation, we have:
3xy^2 * dy/dx - 3x^2y = 0
Substituting the given point:
We are given the point (1, -1). Substituting x = 1 and y = -1 into the equation, we get:
3(1)(-1)^2 * dy/dx - 3(1)^2(-1) = 0
-3dy/dx + 3 = 0
Simplifying the equation, we have:
-3dy/dx = -3
dy/dx = 1
The slope of the tangent line at the point (1, -1) is 1.
Therefore, the correct answer is option A) 1/2.
We are given the equation of a curve: xy^3 - yx^3 = 6. We need to find the slope of the tangent line at the point (1, -1).
Solution:
To find the slope of the tangent line at a given point on the curve, we need to take the derivative of the curve equation with respect to x and then substitute the x-coordinate of the given point into the derivative equation.
Let's differentiate the equation xy^3 - yx^3 = 6 with respect to x.
Differentiating the equation:
To differentiate the equation, we will use the product rule and the chain rule of differentiation.
The product rule states that if u(x) = f(x)g(x), then the derivative of u(x) with respect to x is given by:
u'(x) = f'(x)g(x) + f(x)g'(x)
The chain rule states that if y = f(u) and u = g(x), then the derivative of y with respect to x is given by:
dy/dx = dy/du * du/dx
Differentiating xy^3 - yx^3 = 6 with respect to x using the product rule, we get:
y^3 + 3xy^2 * dy/dx - 3x^2y - y^3 * dx/dx = 0
Simplifying the equation, we have:
3xy^2 * dy/dx - 3x^2y = 0
Substituting the given point:
We are given the point (1, -1). Substituting x = 1 and y = -1 into the equation, we get:
3(1)(-1)^2 * dy/dx - 3(1)^2(-1) = 0
-3dy/dx + 3 = 0
Simplifying the equation, we have:
-3dy/dx = -3
dy/dx = 1
The slope of the tangent line at the point (1, -1) is 1.
Therefore, the correct answer is option A) 1/2.
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Community Answer
For the curve xy3- yx3= 6, the slope of the tangent line at the point ...
Concept:
Let y = f(x) be the equation of a curve, then the slope of the tangent at any point say (x1, y1) is given by:
Calculation:
Given curve is xy3 - yx3 = 6
Now by partially differentiating the equation of curve with respect to x we get;
The slope(m) i.e. dy/dx of the tangent at (1, -1) is:
m = -1
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For the curve xy3- yx3= 6, the slope of the tangent line at the point (1, -1) is:a)-1b)1/2c)2d)1Correct answer is option 'A'. Can you explain this answer?
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