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Jodis
Questions (3)
The curve y=x^3-3x^2-8x+4 has tangent L at point P (-1,8). Given that the Line M is parallel to L and is also a tangent to Q
4 answers
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The curve y=x^3-3x^2-8x+4 has tangent L at point P (1,-8). Given that the Line M is parallel to L and is also a tangent to Q
1 answer
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Please: Differential equation
(dy/dx)^2 + sin(x)cos^2(x)dy/dx - sin^4(x)=0 Answer is y=cos(x)+c or y=1/3 cos^3(x) - cos(x) + c
4 answers
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Answers (4)
Thanks, I calculated both tangents but was stuck on the next part. Thanks for explanation
I found the x for Q to be 3 and subbing x into the equation found y-coordinate as -20 Q(3,-20) P(-1,8) However using the distance equation I get 20root2 not 16root2. Can you check whether my working is right?
So for line 3... can I use (sin(x) (1+sin^2x)/2)^2 instead of sin^2x(1+sin^2x)^2/4 and then root the answer to get ±sin(x)(1+sin^2x)/2 Thanks again
Thanks I got cos(x)+c which matches my working. However, can you explain lines 3 and 4 i.e. sin^2x(1+sin^2x)^2/4 ---> ±sinx(1+sin^2x)/2
