What is the slope of the tangent line to the unit circle at the point (- squarer

Question

What is the slope of the tangent line to the unit circle at the point (- squareroot 2/2, squareroot 2/2)? Find the area of the region that is bounded by the given curve and lies in the specified sector. r = e^- theta/4, pi/2 lessthanorequalto theta lessthanorequalto pi. Consider the system of parametric equations x = t^2 - 2t, y = t+ 1. (a) Does this parametric curve have any horizontal tangents? (b) Does this parametric curve have any vertical tangents? (c) What is the area under this parametric curve on the interval 1 lessthanorequalto t lessthanorequalto 3?

Explanation / Answer

16) we have the unit circle with radius r=1 is x^2+y^2=1 and given point (-sqrt(2)/2,sqrt(2)/2)

differentiating unit circle with respect to x

2x+2y(dy/dx)=0

2y(dy/dx)=-2x

y(dy/dx)=-x

dy/dx=-x/y

at point (-sqrt(2)/2,sqrt(2)/2) the slope m=dy/dx=-x/y

m=(sqrt(2)/2)/(sqrt(2)/2)=1

m=1

the slope of tangent line to the unit circle at (-sqrt(2)/2,sqrt(2)/2) is 1

17) We have given r=e^(-/4),pi/2<=<=pi

we know the Area bounded by polar curve is 1/2 * integration of (x=a to b) (r^2)d

A=1/2 * integration of (x=pi/2 to pi) ((e^(-/4))^2)d

=1/2 * integration of (x=pi/2 to pi) (e^(-/2))d

=1/2*[e^(-/2)/(-1/2)] from x=pi/2 to pi since integration of e^(ax)dx=e^(ax)/a

=1/2*[(-2)*e^(-/2)] from x=pi/2 to pi

=1/2*[(-2)*e^(-pi/2)-((-2)*e^(-pi/4))]

=[-e^(-pi/2)+e^(-pi/4)]

=0.248

Area bounded by polar curve is 0.248

18) We have given x=t^2-2t,y=t+1

a) dx/dt=2t-2,dy/dt=1

dy/dx =(dy/dt)/(dx/dt) =1/(2t-2)

you will have horizontal tangents when the numerator is 0 but in our case numerator is doesn't 0 and 2t-2 not equal to 0

so the given curve does not have horizontal tangents

b) you will have vertical tangents when the denominator is 0 and numerator is not equal to 0

(2t-2)=0 implies t=1

vertical tangent is t=1

c) We have given x=t^2-2t,y=t+1 and 1<=t<=3

Let f(t)=x=t^2-2t and g(t)=y=t+1

f'(t)=2t-2

we know the formula for area under the parametric curve is integration of (t=1 to 3) (g(t)f'(t)dt)

Area A=integration of (t=1 to 3) ((t+1)(2t-2)dt)

=integration of (t=1 to 3)(2t^2-2t+2t-2)dt

=integration of (t=1 to 3)(2t^2-2)dt

=[(2/3)t^3-2t] from t=1 to 3

=[((2/3)*27-6)-((2/3)-2)]

=[18-6-2/3+2]

=[14-2/3]

=40/3

A=40/3

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