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13 Nov 2019

find an equation in x and y for the line tangent to the curve at the given number.

4. a) x(t)=t^3 ; y(t)=(2t+1)^3, t=1

b) x(t)= sin^3(t) ; y(t)=2cos(t), t= pi/4

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Trinidad Tremblay
Trinidad TremblayLv2
11 Apr 2019

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Related questions

Please explain all the steps as well as the sketches. Thanks

Show transcribed image text

Please answer the following with detailed explainations.

1). Find the slopes of the tangent lines to the given curves at the indicated points.

x= t^3 - t, y=t^4 - 5t^2 + 4, a) t= -1, b) t= 1, c) (0,4)

2) Find the slopes of the tangent lines to the given curves at the indicated points.

x= cos 2t, y= sin 3t, a) t= pi/2, b) t= 3pi/2, c) (1,0)

3) Sketch the graph and find the slope of the curve at the given point.

x=t^3 - t, y= t^4 - 5t^2 + 4, at (0,0)

4) Identify all points at which the cuve has a) horizontal tangent and b) a vertical tangent.

x= cos 2t, y =sin 7t

5) Find the area enclosed by the given curve.

x= tsin t, y= tcos t, from -pi/2 <= t <= pi/2

Thank you!

Please answer the following THREE questions in detail.

3) Sketch the graph and find the slope of the curve at the given point.

x=t^3 - t, y= t^4 - 5t^2 + 4, at (0,0)

4) Identify all points at which the cuve has a) horizontal tangent and b) a vertical tangent.

x= cos 2t, y =sin 7t

5) Find the area enclosed by the given curve.

x= tsin t, y= tcos t, from -pi/2 <= t <= pi/2

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