Find the slope of the line tangent to the curve 3x2y+cosy=3√+π2 at the point (−1,π6) .
To find the slope of the line tangent to the curve, we need to find the derivative of the curve at the point (-1, π/6).
First, we rearrange the equation of the curve to isolate y:
3x^2y + cos(y) = 3√x + π/2
3x^2y = -cos(y) + 3√x + π/2
y = (-cos(y) + 3√x + π/2) / (3x^2)
Now, we can find the derivative of y with respect to x using implicit differentiation.
Taking the derivative of both sides of the equation with respect to x:
d/dx [y] = d/dx [(-cos(y) + 3√x + π/2) / (3x^2)]
Using the quotient rule on the right side:
dy/dx = [(-sin(y)dy/dx) + (3/(2√x))]/(3x^2) - (2(-cos(y) + 3√x + π/2))/(3x^3)
At the point (-1, π/6), we substitute x=-1 and y=π/6 into the derivative:
dy/dx = [(-sin(π/6)dy/dx) + (3/(2√(-1)))]/((-1)^2) - (2(-cos(π/6) + 3√(-1) + π/2))/((-1)^3)
dy/dx = [(-1/2)dy/dx + (3/(2i))] - (2(-√3/2 + 3i + π/2))/(-1)
dy/dx = (-1/2)dy/dx + (3/(2i)) + √3/2 - 3i - π/2
Now, we rearrange the equation to solve for dy/dx (the slope):
(1/2)dy/dx = (3/(2i)) + √3/2 - 3i - π/2
dy/dx = 2[(3/(2i)) + √3/2 - 3i - π/2]
dy/dx = (3i - 3√3 + 4i - 2π)/i
dy/dx = (3i + 4i - 3√3 - 2π)/i
dy/dx = (7i - 3√3 - 2π)/i
The slope of the line tangent to the curve at the point (-1, π/6) is given by the imaginary part of the expression (7i - 3√3 - 2π)/i:
slope = -3√3 - 2π
To find the slope of the line tangent to the curve at a specific point, we can use implicit differentiation.
First, let's differentiate both sides of the equation with respect to x to find the derivative.
For the term 3x^2y, we use the product rule: d/dx (xy) = y + x(dy/dx).
For the term cos(y), we use the chain rule: d/dx (cos(y)) = -sin(y) * (dy/dx).
For the term 3√(x) + π/2, the derivative of √(x) with respect to x is 1/(2√(x)).
So, applying these rules, we get:
6xy + 3x^2(dy/dx) - sin(y)(dy/dx) * dy/dx + 1/(2√(x)) = 0.
Next, let's substitute the values of x and y from the given point (-1, π/6) into the equation.
When x = -1 and y = π/6, we have:
6(-1)(π/6) + 3(-1)^2(dy/dx) - sin(π/6)(dy/dx) * dy/dx + 1/(2√(-1)) = 0.
Simplifying this equation will give us the value of dy/dx, which represents the slope of the tangent line at the point (-1, π/6).
To find the slope of the line tangent to a curve at a specific point, we need to differentiate the equation of the curve with respect to x.
Let's differentiate the given equation, 3x^2y + cos(y) = 3√(x) + π/2, with respect to x using the chain rule:
d/dx [3x^2y + cos(y)] = d/dx [3√(x) + π/2]
To differentiate the left side of the equation, we need to use the product rule. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by:
d/dx [u(x) * v(x)] = u'(x) * v(x) + u(x) * v'(x)
Applying the product rule to the left side of the equation, we have:
[3x^2 * dy/dx] + [6xy * y' + (-sin(y) * dy/dx)] = [d/dx (3√(x) + π/2)]
Now, let's solve the equation for dy/dx, the derivative of y with respect to x:
3x^2 * dy/dx + 6xy * y' - sin(y) * dy/dx = d/dx (3√(x) + π/2)
Rearranging the terms and factoring out dy/dx:
dy/dx (3x^2 - sin(y)) = d/dx (3√(x) + π/2) - 6xy * y'
Dividing both sides by (3x^2 - sin(y)), we get:
dy/dx = [d/dx (3√(x) + π/2) - 6xy * y'] / (3x^2 - sin(y))
Now, plugging in the values of x = -1 and y = π/6 into the equation, we can find the slope of the tangent line at the point (-1, π/6):
dy/dx = [d/dx (3√(-1) + π/2) - 6(-1)(π/6) * y'] / (3(-1)^2 - sin(π/6))
Calculating the derivatives, we have:
dy/dx = [d/dx (3) - (π/3) * y'] / (3 - sin(π/6))
The derivative of a constant 3 is 0, and the derivative of y' (the derivative of y) is the second derivative of y with respect to x, which we don't know. However, at the point (-1, π/6), we only need to find the value of dy/dx, not y' itself.
Thus, the value of dy/dx, which represents the slope of the tangent line at the given point, will be:
dy/dx = (0 - (π/3) * y') / (3 - sin(π/6))
To calculate this slope, we need to know the value of y' (the second derivative of y with respect to x), which is not given in the question.