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The equation of the tangent to the curve y=2 sin x+sin 2x at x=π/3 is equal to
- a)2y=3√3
- b)y=3√3
- c)2y+3√3=0
- d)y+3√3=0
Correct answer is option 'A'. Can you explain this answer?
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The equation of the tangent to the curve y=2 sin x+sin 2x at x=π/3 ...
To find the equation of the tangent to the curve y=2 sin x sin 2x at a specific value of x, we need to find the slope of the curve at that point.
Using the product rule and chain rule, we can find the derivative of y with respect to x:
y' = 2(cos x sin 2x + 2sin x cos 2x)
Now we can plug in the given value of x to find the slope of the curve at that point:
y'(x) = 2(cos x sin 2x + 2sin x cos 2x)
y'(π/4) = 2(cos(π/4) sin 2(π/4) + 2sin(π/4) cos 2(π/4))
y'(π/4) = 2(√2/2 × √2/2 + 2 × √2/2 × 0))
y'(π/4) = √2
So the slope of the tangent to the curve at x=π/4 is √2.
To find the equation of the tangent, we also need a point on the tangent. We know that the point (π/4, 2 sin(π/4) sin 2(π/4)) lies on both the curve and the tangent.
The y-coordinate of this point can be simplified:
2 sin(π/4) sin 2(π/4) = 2 × √2/2 × √2/2
= 1
So the point (π/4, 1) lies on the tangent.
Using the point-slope form of the equation of a line, we can write the equation of the tangent:
y - 1 = √2(x - π/4)
y = √2x - (π/4)√2 + 1
This is the equation of the tangent to the curve y=2 sin x sin 2x at x=π/4.
Using the product rule and chain rule, we can find the derivative of y with respect to x:
y' = 2(cos x sin 2x + 2sin x cos 2x)
Now we can plug in the given value of x to find the slope of the curve at that point:
y'(x) = 2(cos x sin 2x + 2sin x cos 2x)
y'(π/4) = 2(cos(π/4) sin 2(π/4) + 2sin(π/4) cos 2(π/4))
y'(π/4) = 2(√2/2 × √2/2 + 2 × √2/2 × 0))
y'(π/4) = √2
So the slope of the tangent to the curve at x=π/4 is √2.
To find the equation of the tangent, we also need a point on the tangent. We know that the point (π/4, 2 sin(π/4) sin 2(π/4)) lies on both the curve and the tangent.
The y-coordinate of this point can be simplified:
2 sin(π/4) sin 2(π/4) = 2 × √2/2 × √2/2
= 1
So the point (π/4, 1) lies on the tangent.
Using the point-slope form of the equation of a line, we can write the equation of the tangent:
y - 1 = √2(x - π/4)
y = √2x - (π/4)√2 + 1
This is the equation of the tangent to the curve y=2 sin x sin 2x at x=π/4.
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The equation of the tangent to the curve y=2 sin x+sin 2x at x=π/3 is equal toa)2y=3√3b)y=3√3c)2y+3√3=0d)y+3√3=0Correct answer is option 'A'. Can you explain this answer?
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