Please, I need some help with this one.
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
y sin 12x = x cos 2y, (π/2, π/4)
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
y sin 12x = x cos 2y, (π/2, π/4)
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Let y be related to x, such that:
y sin 12x = x cos 2y
We want to rewrite it in a form:
dy/dx = f(x,y) ... f being some function of x and y.
So, beginning the implicit differentiation:
y sin 12x = x cos 2y
D[ y sin 12x ]/dx = D[ x cos 2y ]/dx ... Differentiate both sides with respect to x.
y(12)(cos 12x) + (y')sin 12x = x(2)(-sin 2y) + (1)cos 2y
Abiding the chain rule and the product rule.
12y cos 12x + (y') sin 12x = cos 2y - 2x sin 2y
(y') sin 12x = cos 2y - 2x sin 2y - 12y cos 12x
y' = (cos 2y - 2x sin 2y - 12y cos 12x) / sin 12x
dy/dx = (cos 2y - 2x sin 2y - 12y cos 12x) / sin 12x
Since dy/dx is the derivative of the curve, it represents the slope of the tangent line to the function. So all we need to do is plug in the coordinates of a point:
Let M be the slope of the tangent line to the curve of x and y:
M(x,y) = (cos 2y - 2x sin 2y - 12y cos 12x) / sin 12x
M(π/2, π/4) = (cos 2[π/2] - 2[π/2] sin 2[π/4] - 12[π/4] cos 12[π/2]) / sin 12[π/2]
M(π/2, π/4) = (cos π - 2 sin π/2 - 12(π/4) cos 6π) / sin 6π
M(π/2, π/4) = (-1 - 2(-1) - 12(π/4) cos 6π) / sin 6π
M(π/2, π/4) = (1 - 12(π/4)(1)) / sin 6π
M(π/2, π/4) = undefined
It took me this long to realize than sin 6π = 0.
M(π/2, π/4) does not exist.
For this particular problem, the derivative of the curve at the point does not exist, which (I hope, if my work was done right) means that the tangent line is just a vertical line at
x = π/2.
Tangent line at (π/2, π/4):
x = π/2
~~~
y sin 12x = x cos 2y
We want to rewrite it in a form:
dy/dx = f(x,y) ... f being some function of x and y.
So, beginning the implicit differentiation:
y sin 12x = x cos 2y
D[ y sin 12x ]/dx = D[ x cos 2y ]/dx ... Differentiate both sides with respect to x.
y(12)(cos 12x) + (y')sin 12x = x(2)(-sin 2y) + (1)cos 2y
Abiding the chain rule and the product rule.
12y cos 12x + (y') sin 12x = cos 2y - 2x sin 2y
(y') sin 12x = cos 2y - 2x sin 2y - 12y cos 12x
y' = (cos 2y - 2x sin 2y - 12y cos 12x) / sin 12x
dy/dx = (cos 2y - 2x sin 2y - 12y cos 12x) / sin 12x
Since dy/dx is the derivative of the curve, it represents the slope of the tangent line to the function. So all we need to do is plug in the coordinates of a point:
Let M be the slope of the tangent line to the curve of x and y:
M(x,y) = (cos 2y - 2x sin 2y - 12y cos 12x) / sin 12x
M(π/2, π/4) = (cos 2[π/2] - 2[π/2] sin 2[π/4] - 12[π/4] cos 12[π/2]) / sin 12[π/2]
M(π/2, π/4) = (cos π - 2 sin π/2 - 12(π/4) cos 6π) / sin 6π
M(π/2, π/4) = (-1 - 2(-1) - 12(π/4) cos 6π) / sin 6π
M(π/2, π/4) = (1 - 12(π/4)(1)) / sin 6π
M(π/2, π/4) = undefined
It took me this long to realize than sin 6π = 0.
M(π/2, π/4) does not exist.
For this particular problem, the derivative of the curve at the point does not exist, which (I hope, if my work was done right) means that the tangent line is just a vertical line at
x = π/2.
Tangent line at (π/2, π/4):
x = π/2
~~~
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Take the partial derivative with respect to x. Then take the partial derivative with respect to y. Plug in (pi/2,pi/4). plug it into the equation for a tangent line.