What is the slope of the polar curve #f(theta) = theta - sec^3theta+thetasin^3theta # at #theta = (5pi)/8#?

Answer 1

#dy/dx=-0.54#

For a polar function #f(theta)#, #dy/dx=(f'(theta)sintheta+f(theta)costheta)/(f'(theta)costheta-f(theta)sintheta)#
#f(theta)=theta-sec^3theta+thetasin^3theta#
#f'(theta)=1-3(sec^2theta)(d/dx[sectheta])-sin^3theta+3thetasin^2theta(d/dx[sintheta])#
#f'(theta)=1-3sec^3thetatantheta-sin^3theta+3thetasin^2thetacostheta#
#f'((5pi)/3)=1-3sec^3((5pi)/3)tan((5pi)/3)-sin^3((5pi)/3)+3((5pi)/3)sin^2((5pi)/3)cos((5pi)/3)~~-9.98#
#f((5pi)/3)=((5pi)/3)-sec^3((5pi)/3)+((5pi)/3)sin^3((5pi)/3)~~-6.16#
#dy/dx=(-9.98sin((5pi)/3)-6.16cos((5pi)/3))/(-9.98cos((5pi)/3)+6.16sin((5pi)/3))=-0.54#
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Answer 2

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Then, substitute (To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\thetaTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \thetaTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta =To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\fracTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dxTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\piTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{dTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\thetaTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}}To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} \To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} \To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} )To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ).To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) intoTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). HereTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into theTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, (To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expressionTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( yTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression toTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y =To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to findTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = fTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find theTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slopeTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\thetaTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope atTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta)To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope at thatTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta) \To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope at that pointTo find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta) )To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope at that point.To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta) ) andTo find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope at that point.To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta) ) and (To find the slope of the polar curve ( f(\theta) = \theta - \sec^3(\theta) + \theta \sin^3(\theta) ) at ( \theta = \frac{5\pi}{8} ), first, calculate the derivative of ( f(\theta) ) with respect to ( \theta ) using the polar coordinate formula for derivatives. Then, substitute ( \theta = \frac{5\pi}{8} ) into the derivative expression to find the slope at that point.To find the slope of the polar curve at a given point, ( \theta = \frac{5\pi}{8} ), you'll need to find ( \frac{{dy}}{{d\theta}} ) and ( \frac{{dx}}{{d\theta}} ) and then calculate ( \frac{{dy}}{{dx}} ) using the formula ( \frac{{\frac{{dy}}{{d\theta}}}}{{\frac{{dx}}{{d\theta}}} ). Here, ( y = f(\theta) ) and ( x = \theta ). So, ( \frac{{dy}}{{d\theta}} ) is the derivative of ( f(\theta) ) with respect to ( \theta ), and ( \frac{{dx}}{{d\theta}} ) is simply 1. After finding these derivatives, plug in ( \theta = \frac{5\pi}{8} ) to get ( \frac{{dy}}{{dx}} ), which gives you the slope of the curve at that point.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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