Steps for finding the tangent line to a polar curve at a specific point

We’ll find the equation of the tangent line to a polar curve in much the same way that we discover the tangent line to a cartesian curve.

We’ll follow these steps:


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1. Find the slope of the tangent line ???m???, using the formula

???m = fracdydx = frac fracdrd hetasin heta+rcos heta fracdrd hetacos heta-rsin heta???

remembering to plug the worth of??? heta??? at the tangent allude into???dy/dx??? to acquire a real-number value for the slope ???m???.

You are watching: Slope of tangent line to polar curve

2. Find???x_1??? and???x2??? by plugging the value of??? heta??? at the tangent point right into the convariation formulas

???x=rcos heta???

???y=rsin heta???

3. Plug the slope???m??? and the point???(x_1,y_1)??? into the point-slope formula for the equation of a line

???y-y_1=m(x-x_1)???


How to uncover the equation of the tangent line



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Building the tangent line equation step-by-step

Example

Find the tangent line to the polar curve at the provided suggest.

???r=1+2cos heta???

at ??? heta=fracpi4???

We’ll start by calculating ???dr/d heta???, the derivative of the provided polar equation, so that we can plug it right into the formula for the slope of the tangent line.

???r=1+2cos heta???

???fracdrd heta=-2sin heta???

Plugging ???dr/d heta??? and also the offered polar equation ???r=1+2cos heta??? into the formula for ???dy/dx???, we get

???m=fracdydx=frac(-2sin heta)sin heta+(1+2cos heta)cos heta(-2sin heta)cos heta-(1+2cos heta)sin heta???

???m=fracdydx=frac-2sin^2 heta+cos heta+2cos^2 heta-2sin hetacos heta-sin heta-2sin hetacos heta???

???m=fracdydx=frac-2sin^2 heta+cos heta+2cos^2 heta-4sin hetacos heta-sin heta???

Plugging the worth of ??? heta=pi/4??? right into the slope equation, we’ll gain a real-number worth for the slope ???m???.

???m=fracdydx=frac-2sin^2fracpi4+cosfracpi4+2cos^2fracpi4-4sinfracpi4cosfracpi4-sinfracpi4???

???m=fracdydx=frac-2left(fracsqrt22 ight)^2+fracsqrt22+2left(fracsqrt22 ight)^2-4cdotfracsqrt22cdotfracsqrt22-fracsqrt22???

???m=fracdydx=frac-2left(frac24 ight)+fracsqrt22+2left(frac24 ight)-4cdotfrac24-fracsqrt22???

???m=fracdydx=frac-1+fracsqrt22+1-2-fracsqrt22???

???m=fracdydx=fracfracsqrt22-frac42-fracsqrt22???

???m=fracdydx=fracfracsqrt22frac-4-sqrt22???

???m=fracdydx=fracsqrt22left(frac2-4-sqrt2 ight)???

???m=fracdydx=fracsqrt2-4-sqrt2???

If we desire to remove the square root in the denominator, we have the right to multiply by the conjugate.

???m=fracdydx=fracsqrt2-4-sqrt2left(frac-4+sqrt2-4+sqrt2 ight)???

???m=fracdydx=frac-4sqrt2+216-2???

???m=fracdydx=frac-4sqrt2+214???

???m=fracdydx=frac-2sqrt2+17???

???m=fracdydx=frac1-2sqrt27???


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Now we desire to discover ???x_1??? and also ???y_1??? by plugging the worth of ??? heta??? at the tangent suggest and the offered polar equation ???r=1+2cos heta??? into the conversion formulas

???x=rcos heta???

???x_1=left(1+2cosfracpi4 ight)cosfracpi4???

???x_1=left<1+2left(fracsqrt22 ight) ight>fracsqrt22???

???x_1=left(1+sqrt2 ight)fracsqrt22???

???x_1=fracsqrt2+22???

???x_1=frac2+sqrt22???

and

???y=rsin heta???

???y_1=left(1+2cosfracpi4 ight)sinfracpi4???

???y_1=left<1+2left(fracsqrt22 ight) ight>fracsqrt22???

???y_1=left(1+sqrt2 ight)fracsqrt22???

???y_1=fracsqrt2+22???

???y_1=frac2+sqrt22???

Plugging ???m??? and ???left(x_1,y_1 ight)??? into the point-slope formula for the equation of a line, we get

???y-y_1=m(x-x_1)???

???y-frac2+sqrt22=frac1-2sqrt27left(x-frac2+sqrt22 ight)???

???y-frac2+sqrt22=frac1-2sqrt27x-frac2+sqrt2-4sqrt2-414???

???y-frac2+sqrt22=frac1-2sqrt27x-frac-3sqrt2-214???

???y-frac2+sqrt22=frac1-2sqrt27x+frac3sqrt2+214???

???2y-left(2+sqrt2 ight)=frac2-4sqrt27x+frac3sqrt2+27???

Eliminate the fractions by multiplying through by ???7???.

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???14y-7left(2+sqrt2 ight)=left(2-4sqrt2 ight)x+3sqrt2+2???

???14y=left(2-4sqrt2 ight)x+3sqrt2+2+7left(2+sqrt2 ight)???

???14y=left(2-4sqrt2 ight)x+3sqrt2+2+14+7sqrt2???

???14y=left(2-4sqrt2 ight)x+16+10sqrt2???

???14y-left(2-4sqrt2 ight)x=16+10sqrt2???

???7y-left(1-2sqrt2 ight)x=8+5sqrt2???

The equation of the tangent line is ???7y-left(1-2sqrt2 ight)x=8+5sqrt2???.