document.write( "Question 1040748: Hello amazing tutors, can you help me solve this question? Thank youu\r
\n" ); document.write( "\n" ); document.write( "The equation of a curve is \"+y+=+sqrt%285x+%2B+4%29+\"\r
\n" ); document.write( "\n" ); document.write( "(i) Calculate the gradient of the curve at the point where x = 1
\n" ); document.write( "(ii) A point with coordinates (x, y) moves along the curve in such a way that the rate of increase of x has the constant value 0.03 units per second. Find the rate of increase of y at the instant when x = 1
\n" ); document.write( "(iii) Find the area enclosed by the curve, the x-axis, the y-axis and the line x = 1
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Algebra.Com's Answer #655641 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
(i) \"y+=+sqrt%285x%2B4%29\" ==> \"dy%2Fdx+=+5%2F%282sqrt%285x%2B4%29%29\" ==> \"dy%281%29%2Fdx+=+5%2F%282sqrt%285%2A1%2B4%29%29+=+5%2F%282%2A3%29+=+5%2F6\"\r
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\n" ); document.write( "\n" ); document.write( "(ii)\"y+=+sqrt%285x%2B4%29\" ==> \"dy%2Fdt+=+%285%2A%28dx%2Fdt%29%29%2F%282sqrt%285x%2B4%29%29\" ==> \"dy%28x=1%29%2Fdt+=+%285%2A0.03%29%2F%282sqrt%285%2A1%2B4%29%29+=+0.075%2F3+=+0.025\" unit per second.\r
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\n" ); document.write( "\n" ); document.write( "(iii) Area = \"int%28sqrt%285x%2B4%29%2C+dx%2C+-4%2F5%2C1%29\"\r
\n" ); document.write( "\n" ); document.write( "= \"%281%2F5%29int%28sqrt%285x%2B4%29%2C+d%285x%2B4%29%2C+-4%2F5%2C1%29\"\r
\n" ); document.write( "\n" ); document.write( "= \"%281%2F5%29%282%2F3%29%28%285x%2B4%29%5E%283%2F2%29%29%5E1%5B-4%2F5%5D\"\r
\n" ); document.write( "\n" ); document.write( "= \"%282%2F15%29%289%5E%283%2F2%29+-+0%5E%283%2F2%29%29+=+%282%2F15%29%2A27+=+18%2F5\"
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