A curve in the xy-plane is defined by the parametric equations #x = t^3 + 2# and #y = t^2 - 5t# how do you find the slope of the line tangent to the curve at the point where x = 10?

Answer 1

The slope of the tangent is #-1/12#

The curve is defined by the parametric equations:

# { (x=t^3+2), (y=t^2-5t) :} #

When #x=10# then:

# x=t^3+2 => 10=t^3+2 #
# :. t^3=8 #
# :. \ \ t=2 #

When #t=2# then:

# y=t^2-5t => y=4-10 = -6 #

So the coordinate of interest with #t=2# is #(10,-6)#

We now need the derivative to establish the gradient of the tangent:

# x=t^3+2 \ => dx/dt=3t^2 #
# y=t^2-5t => dy/dt=2t-5 #

By the chain rule we have:

# dy/dx = (dy/dt)/(dx/dt) #

# " " = (2t-5)/(3t^2) #

So when #t=2# we have:

# dy/dx = -1/12 #

So our tangent has slope #-1/12# and passes through #(10,-6)#, So using the point/slope formula #y-y_1=m(x-x_1)# for s straight line the required equation is:

# y-(-6)=-1/12(x-10) #
# :. y+6=-1/12x+10/12 #
# :. y=-1/12x-31/6 #

We can confirm this graphically:

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Answer 2

To find the slope of the tangent line to the curve at the point where ( x = 10 ), you need to follow these steps:

  1. Find the corresponding value of ( t ) when ( x = 10 ) by solving the equation ( x = t^3 + 2 ) for ( t ).
  2. Once you have the value of ( t ), substitute it into the equation for ( y = t^2 - 5t ) to find the corresponding ( y )-coordinate.
  3. Differentiate both ( x ) and ( y ) with respect to ( t ) to find ( \frac{dy}{dx} ).
  4. Evaluate ( \frac{dy}{dx} ) at the point where ( x = 10 ) by substituting the values of ( t ), ( x ), and ( y ) obtained earlier.
  5. The value you obtain is the slope of the tangent line to the curve at the point where ( x = 10 ).
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Answer from HIX Tutor

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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