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Mathematics 14 Online
OpenStudy (czesc):

Instant fan/medal: Find the exact value of the slope of the line which is tangent to the curve given by the equation r = 1 + cos θ at pi/2=theta.

OpenStudy (faiqraees):

find the gradient first

OpenStudy (faiqraees):

Do you have the expression for gradient?

OpenStudy (czesc):

@FaiqRaees I don't even know what a gradient is unfortunately, my online course is pretty vague. This is my first time working with polar curves :)

OpenStudy (faiqraees):

Do you know differentiation?

OpenStudy (faiqraees):

I am not able to help you if you wont respond.

OpenStudy (czesc):

@FaiqRaees yes I do know differentiation, but I am not sure how complex the problem is yet

OpenStudy (faiqraees):

Okay so to differentiate y First differentiate 1 Whats the answer?

OpenStudy (czesc):

d/dx of 1 = 0?

OpenStudy (faiqraees):

Perfect Now differentiate cos θ

OpenStudy (czesc):

-sin(theta)

OpenStudy (faiqraees):

Very good d/dx y = 0+ (-sin θ) d/dx y = -sin θ

OpenStudy (faiqraees):

Do you understand how we arrived at the concluded gradient?

OpenStudy (czesc):

yes we derived the polar curve

OpenStudy (faiqraees):

Okay now we have to find the gradient at θ=π/2 so substitute θ=π/2 in the equation of gradient and find out the gradient at θ=π/2.

OpenStudy (czesc):

-sin(pi/2)=-1

OpenStudy (faiqraees):

Yes very good. Now we have the gradient of our tangent. All we need is a point which lies on the tangent. Any idea how can we figure out that point?

OpenStudy (czesc):

graph?

OpenStudy (faiqraees):

We can do that but its complicated. What we will do is, we know that tangent touches the curve at θ=π/2. Which means the coordinate at θ=π/2 will lie both on the curve and the tangent. Agree?

OpenStudy (czesc):

Yep :)

OpenStudy (faiqraees):

So substitute θ=π/2 in the equation of curve to find the y coordinate at θ=π/2

OpenStudy (czesc):

1+cos(pi/2)=1

OpenStudy (faiqraees):

so our coordinates are (π/2,1) Now substitute the coordinates in the equation y= -x+c (where c is a constant to be found)

OpenStudy (faiqraees):

Confused?

OpenStudy (czesc):

1=-pi/2 +c, c=1+(pi/2) ? @FaiqRaees sorry for slow response there

OpenStudy (faiqraees):

yes now substitute the value of c in y= -x+c

OpenStudy (czesc):

y=-x+1+pi/2

OpenStudy (faiqraees):

yes correct, there you have the equation of tangent

OpenStudy (czesc):

So how do I go about finding slope now?

OpenStudy (faiqraees):

Oh btw I am really really sorry, I misread the questio, the question was way shorter. After finding the derivative, you just had to plug θ=π/2 and you would have your answer -1

OpenStudy (czesc):

lol it's fine ! Thank you for everything @FaiqRaees

OpenStudy (faiqraees):

dy/dx = -sinθ dy/dx = -sin(π/2) dy/dx = -1 Slope =-1

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