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In this lesson, we will learn how to find the equation of a curve given the function that describes the slope of its tangent.

Q1:

A curve passes through ( 0 , 1 ) and the tangent at its point ( π₯ , π¦ ) has slope 6 π₯ β 8 π₯ + 1 2 . What is the equation of the curve?

Q2:

A curve passes through ( 0 , 1 ) and the tangent at its point ( π₯ , π¦ ) has slope 4 π₯ β 2 π₯ + 9 2 . What is the equation of the curve?

Q3:

Find the equation of the curve that passes through the point ( β 2 , 1 ) given that the gradient of the tangent to the curve is β 1 1 π₯ 2 .

Q4:

The gradient of the tangent to a curve is β 6 π₯ + 6 π₯ s i n c o s . For π₯ β ο 0 , π 3 ο , the curve has a local minimum value of β 4 6 β 2 9 . Find the equation of the curve.

Q5:

The gradient of the tangent to a curve is β π₯ + π₯ s i n c o s . For π₯ β [ 0 , 2 π ] , the curve has a local minimum value of 3 5 β 2 3 9 . Find the equation of the curve.

Q6:

Find the local minimum value of a curve given that its gradient is d d π¦ π₯ = π₯ + 3 π₯ β 1 8 2 and the local maximum value is 21.

Q7:

Find the equation of a curve which passes through the point ( 0 , 0 ) and, for each point ( π , π ) on the curve, the slope of the tangent at that point is β 3 π₯ β π₯ 5 8 9 .

Q8:

The slope at the point ( π₯ , π¦ ) on the graph of a function is β 3 π 6 π₯ . What is π ( β 3 ) , given that π ( β 5 ) = 9 ?

Q9:

The slope at the point ( π₯ , π¦ ) on the graph of a function is β 4 π β 5 π₯ 7 . What is π ( β 3 ) , given that π ( 9 ) = 9 ?

Q10:

The slope at the point ( π₯ , π¦ ) on the graph of a function is d d s i n c o s π¦ π₯ = β 4 π π π₯ + 5 π π π₯ . Find the equation of the curve if it contains the point ( 1 , 2 ) .

Q11:

The slope at the point ( π₯ , π¦ ) on the graph of a function is d d s i n c o s π¦ π₯ = 3 π π π₯ β 7 π π π₯ . Find the equation of the curve if it contains the point ( 1 , 8 ) .

Q12:

Find the equation of the curve given the gradient of the normal to the curve is β 2 π₯ β 2 and the curve passes through the point ( 1 , 6 ) .

Q13:

Find the equation of the curve given the gradient of the normal to the curve is β 8 π₯ + 4 and the curve passes through the point ( 4 , 2 ) .

Q14:

The second derivative of a curve is β 2 7 3 π₯ + 8 s i n . The curve passes through the point οΎ π 6 , β 4 π 3 + π 9 + 6 ο 2 and the gradient of the tangent at this point is β 8 + 4 π 3 . Find the equation of the curve.

Q15:

If the rate of change of the sales in a factory is inversely proportional to time in weeks, and the sales of the factory after 2 weeks and 4 weeks are 118 units and 343 units, respectively, determine the sales of the factory after 8 weeks.

Q16:

Find the equation of the curve given the gradient of the tangent is 5 ο» π₯ 2 ο s i n 2 and the curve passes through the origin.

Q17:

The gradient of the tangent to a curve passing through the point is equal to . Find the equation of the tangent at the point when is equal to 1.

Q18:

A curve passes through ( 1 , 8 ) and the normal at its point ( π₯ , π¦ ) has slope 8 β 9 π₯ . What is the equation of the curve?

Q19:

Given that the slope at ( π₯ , π¦ ) is 3 π 3 π₯ and π ( 0 ) = β 3 , determine π ( β 3 ) .

Q20:

Given that the slope at ( π₯ , π¦ ) is β 4 π 2 π₯ and π ( 0 ) = 1 , determine π ( 4 ) .

Q21:

The slope at the point ( π₯ , π¦ ) on the graph of a function is 6 π + 2 π₯ . What is π ( π₯ ) , given that π ( 5 ) = 1 l n ?

Q22:

A curve passes through the points ο» π 4 , 8 ο and οΌ 3 π 4 , β 6 ο . Find the equation of the curve given the gradient of the tangent to the curve equals β 7 ( π₯ ) c s c 2 .

Q23:

The gradient of the tangent to a curve is d d π¦ π₯ = π₯ β 1 4 π₯ + 4 5 2 where the value of the local maximum is 9. Find the equation of the curve and the value of the local minimum if it exists.

Q24:

The slope at the point ( π₯ , π ( π₯ ) ) on the graph of a function is 4 β 5 π + 4 π₯ . What is π ( 4 π ) if we know that π ( π ) = β 9 ?

Q25:

The slope at the point ( π₯ , π¦ ) on the graph of a function is 5 π₯ β 2 π₯ . Find the equation of the curve if it contains the point ( π , 5 π + 3 ) .

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