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Author Question: Find the Cartesian equation of the straight line tangent to the plane curve given parametrically by ... (Read 73 times)

olgavictoria

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on: May 26, 2021

Question 1

Find the equation of the tangent line to the curve at the given t.
x = 2 cot t,  y = 2 sin2t  at  t =
◦ x + 2y = 4
◦ x - 2y = 0
◦ 2x - y = 3
◦ x - 2y = 4
◦ 2x + y = 5

Question 2

Find the Cartesian equation of the straight line tangent to the plane curve given parametrically by at the point on the curve where t = -1.
◦ y = x
◦ y = -3x
◦ y = 0
◦ 3x - y =0
◦ x + y = 0
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Marked as best answer by olgavictoria on May 26, 2021

Alyson.hiatt@yahoo.com

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Reply #1 on: May 26, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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olgavictoria

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Reply 2 on: May 26, 2021
:D TYSM

covalentbond

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Reply 3 on: Yesterday
Excellent

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