#dy over dx

37 messages · Page 1 of 1 (latest)

delicate dove
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@trail nebula i assume you know power rule? if so, dy/dx for first one is 4x^3-1, and sub x=1 for gradient of the tangent at that point so the gradient is 4(1)^3-1=3, and you can find a point it passes through by subbing x=1 into original eq so 1^4-1+1=1, so the point is (1,1)

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so y-y1=m(x-x1) for eq of the tangent so y-1=3(x-1)

trail nebula
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Ohh thankss

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I forgot to use the straight line eqatuon

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And stopped after derfintion

delicate dove
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for second one you pretty much do the same, find the gradient of the tangent through differentiation, however as its the normal it is perpendicular to the tangent so you use the negative reciprocal of the gradient of the tangent

trail nebula
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Thankss

delicate dove
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np

trail nebula
delicate dove
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because -x differentiates to -1

trail nebula
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Ohh I thought like -1 was joined onto the power

delicate dove
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as its -x^1, so multiply by power, and then reduce the power by 1, so -1^0

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ahh nah its seperate

trail nebula
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Oh wait I think I get it

delicate dove
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so for the equation of the line you need gradient and a point it passes through

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the question tells you it passes through x=1, so you can sub x=1 into the original equation to find the y value at that point, so you have a point it passes through

trail nebula
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Ohh

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Okk

trail nebula
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Like I got x + 5y = 33

delicate dove
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so differentiate and you get 2x-1 right?

trail nebula
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Yeah

delicate dove
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so sub x=3 to get the gradient of the tangent at that point

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and you get 2(3)-1=5

trail nebula
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Yep

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I got up to that part

delicate dove
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and since the normal is perpendicular to the tangent use the negative reciprocal of the tangent as the gradient of the normal, and it passes through (3,6)

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so y-6=-1/5(x-3)

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rearrange for y=mx+c by first multiplying by 5, so 5y-30=-x+3

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add x to both sides and 30 to both sides, so 5y+x=33

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and when it touches x axis is when y=0

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so just sub y=0

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so x=33

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so the point is (33,0)

trail nebula
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Thanks 🌚