WhatsApp Image 2019-03-31 at 8.42.19 PM (8).jpeg

# WhatsApp Image 2019-03-31 at 8.42.19 PM (8).jpeg - EX = 1 y...

• 1

This preview shows page 1 out of 1 page.

Unformatted text preview: EX * ( + ) = 1 + + y ( 6 ) = 7 Cos(t) en Comph't the equation of the tangent t = IT The equation of tangent line = 10 - 4, _ slop ofe - dy - 75 in(E ) X - X ax 2 . t _when t = TI - 7\ sin ( I) _- 7 * \ 3 x 3) _- 21 X < 3 R TI = 3 2 TT 41 TT _ 1 < 3 4 TT slope _TO get X , y blug 1 + 15 Y ( II ) _ 7 . 1 = 7 the equation . y - * 21: 4 3 = X - ( 1 + IT ? ) 4 TT |6 Area = S y dx = Sy ( E ) . * (+ ), at a Q * ^) yet X / 6 ) = Cos ( t ), y`) = Sin ( t) Area From IT Area \ = - \ S in (* ) . Sin ( 8) d+ IT - S &in ` ) dib)` - 45 1 - cos ( 2 #) d t` . = \$ _ sin ( 2 6 ) | 0 2 IT IT...
View Full Document

• Spring '08
• KijuwaIthink
• Trigraph, #, OFE

{[ snackBarMessage ]}

###### "Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern