An equation of the tangent line to the curve
at the pointis y =
(point intercept form: y = mx+b)
I know it's easier to fine f(3π) ans f'(3π) and then simplifty the exact value of sin(3π) and cos(3π)...
So I got
f(3π) = 3π((5cos(3π)-9sin(3π))) = -15π
f'(3π) = 3π(-5sin(3π)-9cos(3π))+5cos(3π)-9sin(3π) = -26.99
but I dont know what to do with these values and or if i've gone to correct way of solving this. any help would be wonderful!
Your line will beOriginally Posted by youmuggles
An equation of the tangent line to the curve
at the point
(point intercept form: y = mx+b)
I know it's easier to fine f(3π) ans f'(3π) and then simplifty the exact value of sin(3π) and cos(3π)...
So I got
f(3π) = 3π((5cos(3π)-9sin(3π))) = -15π
f'(3π) = 3π(-5sin(3π)-9cos(3π))+5cos(3π)-9sin(3π) = -26.99
but I dont know what to do with these values and or if i've gone to correct way of solving this. any help would be wonderful!
 = f'\left(\frac{\pi}{3}\right)\left(x-\frac{\pi}{3}\right))
The point on the curve, where the tangent is drawn, is (3π, -15π)An equation of the tangent line to the curve
at the point
(point intercept form: y = mx+b)
I know it's easier to fine f(3π) ans f'(3π) and then simplifty the exact value of sin(3π) and cos(3π)...
So I got
f(3π) = 3π((5cos(3π)-9sin(3π))) = -15π
f'(3π) = 3π(-5sin(3π)-9cos(3π))+5cos(3π)-9sin(3π) = -26.99
but I dont know what to do with these values and or if i've gone to correct way of solving this. any help would be wonderful!
The slope of the tangent is (27π - 5 )
Now the equation of the tangent is
(y-y1) = m(x-x1)
Substitute the values to get the answer.
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